Feedback control of dynamic systems (6th ed) SM

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Solutions Manual 6th Edition

Feedback Control of Dynamic Systems . .

Gene F. Franklin .

J. David Powell .

Abbas Emami-Naeini . . . .

Assisted by: H.K. Aghajan H. Al-Rahmani P. Coulot P. Dankoski S. Everett R. Fuller T. Iwata V. Jones F. Safai L. Kobayashi H-T. Lee E. Thuriyasena M. Matsuoka

Chapter 1

An Overview and Brief History of Feedback Control 1.1

Problems and Solutions

1. Draw a component block diagram for each of the following feedback control systems. (a) The manual steering system of an automobile (b) Drebbel’s incubator (c) The water level controlled by a ‡oat and valve (d) Watt’s steam engine with ‡y-ball governor In each case, indicate the location of the elements listed below and give the units associated with each signal. the process the process output signal the sensor the actuator the actuator output signal The reference signal Notice that in a number of cases the same physical device may perform more than one of these functions. Solution: (a) A manual steering system for an automobile: 101

102CHAPTER 1. AN OVERVIEW AND BRIEF HISTORY OF FEEDBACK CONTROL

(b) Drebbel’s incubator:

(c) Water level regulator:

(d) Fly-ball governor:

1.1. PROBLEMS AND SOLUTIONS

103

(e) Automatic steering of a ship:

(f) A public address system:

2. Identify the physical principles and describe the operation of the thermostat in your home or o¢ ce. Solution: A thermostat is a device for maintaining a temperature constant at a desired value. It is equipped with a temperature sensor which detects deviation from the desired value, determines whether the temperature setting is exceeded or not, and transmits the information to a furnace or air conditioner so that the termperature in the room is brought back

104CHAPTER 1. AN OVERVIEW AND BRIEF HISTORY OF FEEDBACK CONTROL S to ra g e chest

R e f in e r

S creen s H eadbox a n d c le a n e rs

M a c h in e chest

P re sse s

D r y in g s e c tio n

R eel

W ire T h ic k s to c k

CV W h ite w a te r

C o n s is te n c y m e te r

C o n tro lle r

Figure 1.1: A paper making machine From Karl Astrom, (1970, page 192) reprinted with permission.

to the desired setting. Examples: Tubes …lled with liquid mercury are attached to a bimetallic strip which tilt the tube and cause the mercury to slide over electrical contacts. A bimetallic strip consists of two strips of metal bonded together, each of a di¤erent expansion coe¢ cient so that temperature changes bend the metal. In some cases, the bending of bimetallic strips simply cause electrical contacts to open or close directly. In most cases today, temperature is sensed electronically using,for example, a thermistor, a resistor whose resistance changes with temperature. Modern computer-based thermostats are programmable, sense the current from the thermistor and convert that to a digital signal. 3. A machine for making paper is diagrammed in Fig. 1.12. There are two main parameters under feedback control: the density of …bers as controlled by the consistency of the thick stock that ‡ows from the headbox onto the wire, and the moisture content of the …nal product that comes out of the dryers. Stock from the machine chest is diluted by white water returning from under the wire as controlled by a control valve (CV). A meter supplies a reading of the consistency. At the “dry end” of the machine, there is a moisture sensor. Draw a signal graph and identify the seven components listed in Problem 1 for (a) control of consistency (b) control of moisture Solution: (a) Control of paper machine consistency:

M o is tu re m e te r

1.1. PROBLEMS AND SOLUTIONS

105

(b) Control of paper machine moisture:

4. Many variables in the human body are under feedback control. For each of the following controlled variables, draw a graph showing the process being controlled, the sensor that measures the variable, the actuator that causes it to increase and/or decrease, the information path that completes the feedback path, and the disturbances that upset the variable. You may need to consult an encyclopedia or textbook on human physiology for information on this problem.

(a) blood pressure (b) blood sugar concentration (c) heart rate (d) eye-pointing angle (e) eye-pupil diameter Solution: Feedback control in human body:

106CHAPTER 1. AN OVERVIEW AND BRIEF HISTORY OF FEEDBACK CONTROL Variable a) Blood pressure

Sensor -Arterial baroreceptors

b) Blood sugar concentration (Glucose) c) Heart rate

-Pancreas

d) Eye p ointing angle e) Pupil diam eter

-Diastolic volum e sensors -Cardiac sym pathetic nerves -Optic nerve -Im age detection -Ro ds

f ) Blood calcium level

-Parathyroid gland detectors

Actuator -Cardiac output -Arteriolar/venous dilation -Pancreas secreting insulin

Inform ation path -A¤erent nerve …b ers

-Electrical stimulation of sino-atrial no de and cardiac muscle -Extrao cular muscles

-M echanical draw of blo o d from heart -Circulating epinephrine -Cranial innervation

-Pupillary sphincter muscles -Ca from b ones to blo o d -Gastrointestinal absorption

-Autonom ous system - Parathorm one horm one a¤ecting e¤ector sites

-Blo o d ‡ow to pancreas

5. Draw a graph of the components for temperature control in a refrigerator or automobile air-conditioning system. Solution:

This is the simplest possible system. Modern cases include computer control as described in later chapters. 6. Draw a graph of the components for an elevator-position control. Indicate how you would measure the position of the elevator car. Consider a combined coarse and …ne measurement system. What accuracies do you suggest for each sensor? Your system should be able to correct for the fact that in elevators for tall buildings there is signi…cant cable stretch as a function of cab load. Solution: A coarse measurement can be obtained by an electroswitch located before the desired ‡oor level. When touched, the controller reduces the motor speed. A “…ne” sensor can then be used to bring the elevator precisely to the ‡oor level. With a sensor such as the one depicted in the …gure, a linear control loop can be created (as opposed to the on-o¤ type of the coarse control).Accuracy required for the course switch is around 5 cm; for the …ne ‡oor alignment, an accuracy of about 2 mm is desirable to eliminate any noticeable step for those entering or exiting the elevator.

Disturbances -Bleeding -Drugs -Stress,Pain -Diet -Exercise

-Horm one release -Exercise

-Head m ovem ent -M uscle twitch -Ambient light -Drugs -Ca need in b one -Drugs

1.1. PROBLEMS AND SOLUTIONS

107

7. Feedback control requires being able to sense the variable being controlled. Because electrical signals can be transmitted, ampli…ed, and processed easily, often we want to have a sensor whose output is a voltage or current proportional to the variable being measured. Describe a sensor that would give an electrical output proportional to: (a) (b) (c) (d) (e) (f) (g) (h) (i) (j)

temperature pressure liquid level ‡ow of liquid along a pipe (or blood along an artery) force linear position rotational position linear velocity rotational speed translational acceleration torque Solution: Sensors for feedback control systems with electrical output. Examples

(a) Temperature: Thermistor- temperature sensitive resistor with resistance change proportional to temperature; Thermocouple; Thyrister. Modern thermostats are computer controlled and programmable. (b) Pressure: Strain sensitive resistor mounted on a diaphragm which bends due to changing pressure

108CHAPTER 1. AN OVERVIEW AND BRIEF HISTORY OF FEEDBACK CONTROL (c) Liquid level: Float connected to potentiometer. If liquid is conductive the impedance change of a rod immersed in the liquid may indicate the liquid level.

(d) Flow of liquid along a pipe: A turbine actuated by the ‡ow with a magnet to trigger an external counting circuit. Hall e¤ect produces an electronic output in response to magnetic …eld changes. Another way: Measure pressure di¤erence from venturi into pressure sensor as in …gure; Flowmeter. For blood ‡ow, an ultrasound device like a SONAR can be used.

(e) Position. When direct mechanical interaction is possible and for “small” displacements, the same ideas may be used. For example a potentiometer may be used to measure position of a mass in an accelerator (h). However in many cases such as the position of an aircraft, the task is much more complicated and measurement cannot be made directly. Calculation must be carried out based on other measurements, for example optical or electromagnetic direction measurements to several known references (stars,transmitting antennas ...); LVDT for linear, RVDT for rotational. (f) Rotational position. The most common traditional device is a poteniometer. Also common are magnetic machines in shich a rotating magnet produces a variable output based on its angle. (g) Linear velocity. For a vehicle, a RADAR can measure linear velocity. In other cases, a rack-and-pinion can be used to translate linear to rotational motion and an electric motor(tachometer) used to measure the speed. (h) Speed: Any toothed wheel or gear on a rotating part may be used to trigger a magnetic …eld change which can be used to trigger an electrical counting circuit by use of a Hall e¤ect (magnetic to electrical) sensor. The pulses can then be counted over a set time interval to produce angular velocity: Rate gyro; Tachometer

1.1. PROBLEMS AND SOLUTIONS

109

(i) Acceleration: A mass movement restrained by a spring measured by a potentiometer. A piezoelectric material may be used instead (a material that produces electrical current with intensity proportional to acceleration). In modern airbags, an integrated circuit chip contains a tiny lever and ’proof mass’whose motion is measured generating a voltage proportional to acceleration.

(j) Force, torque: A dynamometer based on spring or beam de‡ections, which may be measured by a potentiometer or a strain-gauge. 8. Each of the variables listed in Problem 7 can be brought under feedback control. Describe an actuator that could accept an electrical input and be used to control the variables listed. Give the units of the actuator output signal. Solution: (a) Resistor with voltage applied to it ormercury arc lamp to generate heat for small devices. a furnace for a building.. (b) Pump: Pumping air in or out of a chamberto generate pressure. Else, a ’torque motor’produces force.. (c) Valve and pump: forcing liquid in or out of the container. (d) A valve is nromally used to control ‡ow. (e) Electric motor (f) Electric motor (g) Electric motor (h) Electric motor (i) Translational acceleration is usually controlled by a motor or engine to provide force on the vehicle or other object. (j) Torque motor. In this motor the torque is directly proportional to the input (current).

Chapter 2

Dynamic Models Problems and Solutions for Section 2.1 1. Write the di¤erential equations for the mechanical systems shown in Fig. 2.39. For (a) and (b), state whether you think the system will eventually decay so that it has no motion at all, given that there are non-zero initial conditions for both masses, and give a reason for your answer. Figure 2.39: Mechanical systems

Solution: 2003

2004

CHAPTER 2. DYNAMIC MODELS The key is to draw the Free Body Diagram (FBD) in order to keep the signs right. For (a), to identify the direction of the spring forces on the object, let x2 = 0 and …xed and increase x1 from 0. Then the k1 spring will be stretched producing its spring force to the left and the k2 spring will be compressed producing its spring force to the left also. You can use the same technique on the damper forces and the other mass.

Free body diagram for Problem 2.1(a)

(a) m1 x •1 m2 x •2

= =

k1 x1 b1 x_ 1 k2 (x2 x1 )

k2 (x1 k3 (x2

x2 ) y) b2 x_ 2

There is friction a¤ecting the motion of both masses; therefore the system will decay to zero motion for both masses.

Free body diagram for Problem 2.1(b)

m1 x •1 m2 x •2

= =

k1 x1 k2 (x1 x2 ) k2 (x2 x1 ) k3 x2

b1 x_ 1

Although friction only a¤ects the motion of the left mass directly, continuing motion of the right mass will excite the left mass, and that interaction will continue until all motion damps out.

2005

Figure 2.40: Mechanical system for Problem 2.2

x2

x1 . . b1(x1 - x2) k1x1

. . b1(x1 - x2) m2

m1 k2(x1 - x2)

F

k2(x1 - x2)

Free body diagram for Problem 2.1(c) m1 x •1 m2 x •2

= k1 x1 k2 (x1 x2 ) b1 (x_ 1 x_ 2 ) = F k2 (x2 x1 ) b1 (x_ 2 x_ 1 )

2. Write the di¤erential equations for the mechanical systems shown in Fig. 2.40. State whether you think the system will eventually decay so that it has no motion at all, given that there are non-zero initial conditions for both masses, and give a reason for your answer. Solution: The key is to draw the Free Body Diagram (FBD) in order to keep the signs right. To identify the direction of the spring forces on the left side object, let x2 = 0 and increase x1 from 0. Then the k1 spring on the left will be stretched producing its spring force to the left and the k2 spring will be compressed producing its spring force to the left also. You can use the same technique on the damper forces and the other mass.

x1 . . b2(x1 - x2)

k1x1

x2 . . b2(x1 - x2)

m1

m2

k2(x1 - x2) Free body diagram for Problem 2.2

k1 x2

2006

CHAPTER 2. DYNAMIC MODELS Then the forces are summed on each mass, resulting in

m1 x •1 m2 x •2

= k1 x1 k2 (x1 x2 ) b1 (x_ 1 x_ 2 ) = k2 (x1 x2 ) b1 (x_ 1 x_ 2 ) k1 x2

The relative motion between x1 and x2 will decay to zero due to the damper. However, the two masses will continue oscillating together without decay since there is no friction opposing that motion and no ‡exure of the end springs is all that is required to maintain the oscillation of the two masses.

3. Write the equations of motion for the double-pendulum system shown in Fig. 2.41. Assume the displacement angles of the pendulums are small enough to ensure that the spring is always horizontal. The pendulum rods are taken to be massless, of length l, and the springs are attached 3/4 of the way down.

Figure 2.41: Double pendulum

Solution:

2007

θ1

3 l 4

θ2

k m

m

3 l sin θ1 4

3 l sin θ 2 4

De…ne coordinates If we write the moment equilibrium about the pivot point of the left pendulem from the free body diagram, M=

mgl sin

1

ml2 •1 + mgl sin

3 k l (sin 4 1

+

sin

1

9 2 kl cos 16

1

2 ) cos 1

(sin

3 l = ml2 •1 4

sin

1

2)

=0

Similary we can write the equation of motion for the right pendulem mgl sin

2

3 + k l (sin 4

1

sin

2 ) cos 2

3 l = ml2 •2 4

As we assumed the angles are small, we can approximate using sin 1 1, and cos 2 1. Finally the linearized equations 1 ; sin 2 2 , cos 1 of motion becomes, 9 kl ( 16 9 ml•2 + mg 2 + kl ( 16 ml•1 + mg

Or

1

+

1

2)

=

0

2

1)

=

0

2008

CHAPTER 2. DYNAMIC MODELS

•1 + g l •2 + g l

9 k ( 16 m 9 k ( 2+ 16 m +

1

1

2)

=

0

2

1)

=

0

4. Write the equations of motion of a pendulum consisting of a thin, 4-kg stick of length l suspended from a pivot. How long should the rod be in order for the period to be exactly 2 secs? (The inertia I of a thin stick about an endpoint is 13 ml2 . Assume is small enough that sin = .) Solution: Let’s use Eq. (2.14) M =I ;

l 2

O

θ

mg De…ne coordinates and forces Moment about point O.

MO

=

mg

=

1 2• ml 3

l sin = IO • 2

• + 3g sin = 0 2l

2009 As we assumed

is small, • + 3g = 0 2l

The frequency only depends on the length of the rod !2 =

3g 2l s

T

=

2 =2 !

2l =2 3g

l

=

3g = 1:49 m 2 2

q 2l (a) Compare the formula for the period, T = 2 3g with the well known formula for the period of a point mass hanging with a string with q l length l. T = 2 g.

(b) Important! In general, Eq. (2.14) is valid only when the reference point for the moment and the moment of inertia is the mass center of the body. However, we also can use the formular with a reference point other than mass center when the point of reference is …xed or not accelerating, as was the case here for point O. 5. For the car suspension discussed in Example 2.2, plot the position of the car and the wheel after the car hits a “unit bump” (i.e., r is a unit step) using MATLAB. Assume that m1 = 10 kg, m2 = 350 kg, kw = 500; 000 N=m, ks = 10; 000 N=m. Find the value of b that you would prefer if you were a passenger in the car. Solution: The transfer function of the suspension was given in the example in Eq. (2.12) to be: (a) Y (s) = 4 R(s) s + ( mb1 +

b 3 m2 )s

+

kw b m1 m2 (s ks ks (m +m 1 2

+ +

ks b ) kw 2 m1 )s

+ ( mk1wmb 2 )s +

kw k s m1 m2

:

This transfer function can be put directly into Matlab along with the numerical values as shown below. Note that b is not the damping

2010

CHAPTER 2. DYNAMIC MODELS ratio, but damping. We need to …nd the proper order of magnitude for b, which can be done by trial and error. What passengers feel is the position of the car. Some general requirements for the smooth ride will be, slow response with small overshoot and oscillation. From the …gures, b 3000 would be acceptable. There is too much overshoot for lower values, and the system gets too fast (and harsh) for larger values. % Problem 2.5 clear all, close all m1 = 10; m2 = 350; kw = 500000; ks = 10000; B = [ 1000 2000 3000 4000 ]; t = 0:0.01:2; for i = 1:4 b = B(i); num = kw*b/(m1*m2)*[1 ks/b]; den =[1 (b/m1+b/m2) (ks/m1+ks/m2+kw/m1) (kw*b/(m1*m2) kw*ks/(m1*m2)]; sys=tf(num,den); y = step( sys, t ); subplot(2,2,i); plot( t, y(:,1), ’:’, t, y(:,2), ’-’); legend(’Wheel’,’Car’); ttl = sprintf(’Response with b = %4.1f ’,b ); title(ttl); end

6. Write the equations of motion for a body of mass M suspended from a …xed point by a spring with a constant k. Carefully de…ne where the body’s displacement is zero. Solution: Some care needs to be taken when the spring is suspended vertically in the presence of the gravity. We de…ne x = 0 to be when the spring is unstretched with no mass attached as in (a). The static situation in (b) results from a balance between the gravity force and the spring.

2011

From the free body diagram in (b), the dynamic equation results m• x=

kx

mg:

We can manipulate the equation m• x= so if we replace x using y = x +

k x+

m g ; k

m k g,

m• y = ky m• y + ky = 0 The equilibrium value of x including the e¤ect of gravity is at x = m kg and y represents the motion of the mass about that equilibrium point. An alternate solution method, which is applicable for any problem involving vertical spring motion, is to de…ne the motion to be with respect to the static equilibrium point of the springs including the e¤ect of gravity, and then to proceed as if no gravity was present. In this problem, we would de…ne y to be the motion with respect to the equilibrium point, then the FBD in (c) would result directly in m• y=

ky:

7. Automobile manufacturers are contemplating building active suspension systems. The simplest change is to make shock absorbers with a changeable damping, b(u1 ): It is also possible to make a device to be placed in parallel with the springs that has the ability to supply an equal force, u2; in opposite directions on the wheel axle and the car body.

2012

CHAPTER 2. DYNAMIC MODELS (a) Modify the equations of motion in Example 2.2 to include such control inputs. (b) Is the resulting system linear? (c) Is it possible to use the forcer, u2; to completely replace the springs and shock absorber? Is this a good idea? Solution: (a) The FBD shows the addition of the variable force, u2 ; and shows b as in the FBD of Fig. 2.5, however, here b is a function of the control variable, u1 : The forces below are drawn in the direction that would result from a positive displacement of x.

Free body diagram m1 x • = b (u1 ) (y_ x) _ + ks (y x) kw (x m2 y• = ks (y x) b (u1 ) (y_ x) _ + u2

r)

u2

(b) The system is linear with respect to u2 because it is additive. But b is not constant so the system is non-linear with respect to u1 because the control essentially multiplies a state element. So if we add controllable damping, the system becomes non-linear. (c) It is technically possible. However, it would take very high forces and thus a lot of power and is therefore not done. It is a much better solution to modulate the damping coe¢ cient by changing ori…ce sizes in the shock absorber and/or by changing the spring forces by increasing or decreasing the pressure in air springs. These features are now available on some cars... where the driver chooses between a soft or sti¤ ride. 8. Modify the equation of motion for the cruise control in Example 2.1, Eq(2.4), so that it has a control law; that is, let u = K(vr v); where

vr K

= =

reference speed constant:

2013 This is a ‘proportional’control law where the di¤erence between vr and the actual speed is used as a signal to speed the engine up or slow it down. Put the equations in the standard state-variable form with vr as the input and v as the state. Assume that m = 1000 kg and b = 50 N s= m; and …nd the response for a unit step in vr using MATLAB. Using trial and error, …nd a value of K that you think would result in a control system in which the actual speed converges as quickly as possible to the reference speed with no objectional behavior. Solution: 1 b v= u m m

v_ + substitute in u = K (vr

v)

v_ +

b 1 K v= u= (vr m m m

v)

Rearranging, yields the closed-loop system equations, v_ +

b K K v + v = vr m m m

A block diagram of the scheme is shown below where the car dynamics are depicted by its transfer function from Eq. 2.7.

vr

+ Σ −

1 m b s+ m

u

K

Block diagram The transfer function of the closed-loop system is,

V (s) Vr (s)

=

so that the inputs for Matlab are

s+

K m b K m + m

v

2014

CHAPTER 2. DYNAMIC MODELS

num = den

=

K m [1

b K + ] m m

For K = 100; 500; 1000; 5000 We have,

Step Response From: U(1) 1 0.9 0.8

K=500 K=1000

0.6

To: Y(1)

Amplitude

0.7

K=5000 0.5

K=100 0.4 0.3 0.2 0.1 0 0

5

10

15

20

25

30

35

40

Time (sec.)

Time responses We can see that the larger the K is, the better the performance, with no objectionable behaviour for any of the cases. The fact that increasing K also results in the need for higher acceleration is less obvious from the plot but it will limit how fast K can be in the real situation because the engine has only so much poop. Note also that the error with this scheme gets quite large with the lower values of K. You will …nd out how to eliminate this error in chapter 4 using integral control, which is contained in all cruise control systems in use today. For this problem, a reasonable compromise between speed of response and steady state errors would be K = 1000; where it responds is 5 seconds and the steady state error is 5%.

2015 % Problem 2.8 clear all, close all % data m = 1000; b = 50; k = [ 100 500 1000 5000 ]; % Overlay the step response hold on for i=1:length(k) K=k(i); num =K/m; den = [1 b/m+K/m]; step( num, den); end 9. In many mechanical positioning systems there is ‡exibility between one part of the system and another. An example is shown in Figure 2.6 where there is ‡exibility of the solar panels. Figure 2.42 depicts such a situation, where a force u is applied to the mass M and another mass m is connected to it. The coupling between the objects is often modeled by a spring constant k with a damping coe¢ cient b, although the actual situation is usually much more complicated than this. (a) Write the equations of motion governing this system. (b) Find the transfer function between the control input, u; and the output, y: Figure 2.42: Schematic of a system with ‡exibility

Solution: (a) The FBD for the system is

2016

CHAPTER 2. DYNAMIC MODELS

Free body diagrams which results in the equations m• x = k (x y) b (x_ y) _ M y• = u + k (x y) + b (x_ y) _ or k b x + x_ m m k b x x_ + y• + M M x •+

k y m k y+ M

b y_ m b y_ M

=

0

=

1 u M

(b) If we make Laplace Transform of the equations of motion

s2 X + k X M

k b X + sX m m b sX + s2 Y + M

k Y m k Y + M

b sY m b sY M

=

0

=

1 U M

=

0 U

In matrix form, ms2 + bs + k (bs + k)

(bs + k) M s2 + bs + k

X Y

From Cramer’s Rule, ms2 + bs + k 0 (bs + k) U 2 ms + bs + k (bs + k) (bs + k) M s2 + bs + k

det Y

= det =

Finally,

ms2 + bs + k (ms2 + bs + k) (M s2 + bs + k)

2U

(bs + k)

2017

Y U

= =

ms2 + bs + k 2

(ms2 + bs + k) (M s2 + bs + k) (bs + k) ms2 + bs + k mM s4 + (m + M )bs3 + (M + m)ks2

2018

CHAPTER 2. DYNAMIC MODELS

Problems and Solutions for Section 2.2 10. A …rst step toward a realistic model of an op amp is given by the equations below and shown in Fig. 2.43. 107 [V+ s+1 = i =0

Vout

=

i+

V ]

Figure 2.43: Circuit for Problem 10.

Find the transfer function of the simple ampli…cation circuit shown using this model. Solution: As i = 0, (a) Vin V Rin V =

Vout

= = =

=

V

Vout Rf

Rf Rin Vin + Vout Rin + Rf Rin + Rf

107 [V+ V ] s+1 107 Rf V+ Vin s+1 Rin + Rf 107 s+1

Rin Vout Rin + Rf

Rf Rin Vin + Vout Rin + Rf Rin + Rf

2019

Figure 2.44: Circuit for Problem 11.

R

f 107 Rin +R Vout f = in Vin s + 1 + 107 RinR+R f

11. Show that the op amp connection shown in Fig. 2.44 results in Vo = Vin if the op amp is ideal. Give the transfer function if the op amp has the non-ideal transfer function of Problem 2.10. Solution: Ideal case:

Vin V+ V

= V+ = V = Vout

Non-ideal case: Vin = V+ ; V = Vout but, V+ 6= V instead,

Vout

= =

107 [V+ s+1 107 [Vin s+1

V ] Vout ]

2020

CHAPTER 2. DYNAMIC MODELS so, 107

107 Vout 107 = s+1107 = = Vin s + 1 + 107 s + 107 1 + s+1 12. Show that, with the non-ideal transfer function of Problem 2.10, the op amp connection shown in Fig. 2.45 is unstabl e. Figure 2.45: Circuit for Problem 12.

Solution: Vin = V ; V+ = Vout

Vout

107 [V+ V ] s+1 107 [Vout Vin ] s+1

= =

Vout = Vin

107 s+1 107 s+1

1

=

107 107 = s 1 + 107 s 107

The transfer function has a denominator with s 107 ; and the minus sign means the exponential time function is increasing, which means that it has an unstable root. 13. A common connection for a motor power ampli…er is shown in Fig. 2.46. The idea is to have the motor current follow the input voltage and the connection is called a current ampli…er. Assume that the sense resistor, Rs is very small compared with the feedback resistor, R and …nd the transfer function from Vin to Ia : Also show the transfer function when Rf = 1:

2021 Solution: At node A, Vin 0 Vout 0 VB 0 + + =0 Rin Rf R At node B, with Rs Ia +

(93)

R 0

VB 0 VB + R Rs

=

VB

=

VB

0

(94)

RRs Ia R + Rs Rs Ia

The dynamics of the motor is modeled with negligible inductance as Jm •m + b _ m Jm s + b

= Kt Ia = Kt Ia

(95)

At the output, from Eq. 94. Eq. 95 and the motor equation Va = Ia Ra + Ke s Vo

= Ia Rs + Va = Ia Rs + Ia Ra + Ke

Kt Ia Jm s + b

Substituting this into Eq.93 Vin Ia Rs 1 Kt Ia + =0 + Ia Rs + Ia Ra + Ke Rin Rf Jm s + b R This expression shows that, in the steady state when s ! 0; the current is proportional to the input voltage. If fact, the current ampli…er normally has no feedback from the output voltage, in which case Rf ! 1 and we have simply Ia = Vin

R Rin Rs

14. An op amp connection with feedback to both the negative and the positive terminals is shown in Fig 2.47. If the op amp has the non-ideal transfer function given in Problem 10, give the maximum value possible for the r positive feedback ratio, P = in terms of the negative feedback r+R Rin ratio,N = for the circuit to remain stable. Rin + Rf Solution:

2022

CHAPTER 2. DYNAMIC MODELS

Figure 2.46: Op Amp circuit for Problem 14.

Vin V Vout V + Rin Rf Vout V+ 0 V+ + R r

V

0

=

0

Rf Rin Vin + Vout Rin + Rf Rin + Rf = (1 N ) Vin + N Vout r Vout = P Vout = r+R =

V+

Vout

=

= =

Vout Vin

107 [V+ V ] s+1 107 [P Vout (1 s+1

=

= =

N ) Vin

N Vout ]

107 (1 N ) s+1 107 107 P N 1 s+1 s+1 107 (1 N ) 107 P 107 N (s + 1) 107 (1 N ) s + 1 107 P + 107 N

2023

0 < 1 107 P + 107 N P < N + 10 7 15. Write the dynamic equations and …nd the transfer functions for the circuits shown in Fig. 2.48. (a) passive lead circuit (b) active lead circuit (c) active lag circuit. (d) passive notch circuit Solution: (a) Passive lead circuit With the node at y+, summing currents into that node, we get Vu

Vy R1

+C

d (Vu dt

Vy )

Vy =0 R2

(96)

rearranging a bit, C V_ y +

1 1 + R1 R2

1 Vy = C V_ u + Vu R1

and, taking the Laplace Transform, we get Cs + R11 Vy (s) = Vu (s) Cs + R11 + R12 (b) Active lead circuit

Rf V

R1

R2

Vin

Vout

C Active lead circuit with node marked Vin V 0 V d + + C (0 R2 R1 dt

V)=0

(97)

2024

CHAPTER 2. DYNAMIC MODELS

0 Vout Vin V = R2 Rf

(98)

We need to eliminate V . From Eq. 98, V = Vin +

R2 Vout Rf

Substitute V ’s in Eq. 97. 1 R2

Vin

R2 Vout Rf

Vin

1 Vin + C V_ in = R1

1 R1 1 Rf

Vin +

1+

R2 Vout Rf

R2 R1

R2 _ C V_ in + Vout Rf

Vout + R2 C V_ out

Laplace Transform Vout Vin

=

Cs + 1 Rf

1 R1

R2 Cs + 1 +

R2 R1

1

=

s + R1 C Rf R2 s + R11C + R21C

We can see that the pole is at the left side of the zero, which means a lead compensator. (c) active lag circuit

V

R2

R1

C

Rin

Vin

Vout

Active lag circuit with node marked Vin 0 0 V V Vout d = = + C (V Rin R2 R1 dt R2 V = Vin Rin

Vout )

=0

2025 Vin Rin

= 1 Rin

R2 Rin Vin

=

1 R1

1+

R2 R1

Vout

R1 R2 Vin Rin Vin +

+C

d dt

Vout

R2 Vin Vout Rin R2 _ +C Vin V_ out Rin

1 R2 C V_ in = Rin

1 Vout R1

C V_ out

R

Vout Vin

R1 R2 Cs + 1 + R21 Rin R1 Cs + 1 1 1 R2 s + R2 C + R1 C

= =

Rin

s+

1 R1 C

We can see that the pole is at the right side of the zero, which means a lag compensator. (d) notch circuit

V1

C

C

V2

+

+

R

R

R/2

Vin

Vout

2C

−

− Passive notch …lter with nodes marked

C

d 0 V1 d (Vin V1 ) + + C (Vout V1 ) = 0 dt R=2 dt Vin V2 d Vout V2 + 2C (0 V2 ) + = 0 R dt R V2 Vout d = 0 C (V1 Vout ) + dt R

We need to eliminat V1 ; V2 from three equations and …nd the relation between Vin and Vout V1

=

V2

=

Cs 2 Cs +

1 R

1 R

2 Cs +

1 R

(Vin + Vout ) (Vin + Vout )

2026

CHAPTER 2. DYNAMIC MODELS

CsV1 = Cs =

1 Vout R 1 1 R (Vin + Vout ) + R 2 Cs +

CsVout +

Cs 2 Cs +

1 R

1 V2 R

1 R

(Vin + Vout )

Cs +

0

C 2 s2 + R12 Vin 2 Cs + R1

=

"

C 2 s2 + R12 2 Cs + R1

1 Cs + R

#

Vout

C 2 s2 + R12

Vout Vin

1 2(Cs+ R )

= Cs + =

2 Cs + = =

C 2 s2 + R12

1 R

1 2(Cs+ R )

C 2 s2 +

1 R2

1 2 R

C 2 s2 +

1 R2

1 R2 C 2 1 C 2 s2 + 4 Cs R + R2 s2 + R21C 2 4 s2 + RC s + R21C 2

C 2 s2 +

16. The very ‡exible circuit shown in Fig. 2.49 is called a biquad because its transfer function can be made to be the ratio of two second-order or quadratic polynomials. By selecting di¤erent values for Ra ; Rb ; Rc ; and Rd the circuit can realise a low-pass, band-pass, high-pass, or band-reject (notch) …lter. (a) Show that if Ra = R; and Rb = Rc = Rd = 1; the transfer function from Vin to Vout can be written as the low-pass …lter Vout A = 2 Vin s s +2 +1 ! 2n !n where A = !n

= =

R R1 1 RC R 2R2

(99)

1 R

Vout

2027

Figure 2.47: Op-amp biquad

(b) Using the MATLAB comand step compute and plot on the same graph the step responses for the biquad of Fig. 2.43 for A = 1; ! n = 1; and = 0:1; 0:5; and 1:0: Solution: Before going in to the speci…c problem, let’s …nd the general form of the transfer function for the circuit.

Vin V3 + R1 R V1 R V3 V3 V2 V1 Vin + + + Ra Rb Rc Rd

V1 + C V_ 1 R2

= =

C V_ 2

=

V2 Vout R

=

There are a couple of methods to …nd the transfer function from Vin to Vout with set of equations but for this problem, we will directly solve for the values we want along with the Laplace Transform. From the …rst three equations, slove for V1; V2 .

V3 Vin + R1 R V1 R V3

1 + Cs V1 R2

= =

CsV2

=

V2

2028

CHAPTER 2. DYNAMIC MODELS

1 1 + Cs V1 V2 R2 R 1 V1 + CsV2 R 1 R2

V1 V2

1 R

+ Cs 1 R

1

=

1 R2

+ Cs Cs +

C 2 s2 +

0

1 R1 Vin

=

0

Cs

1

=

=

V1 V2

Cs

1 Vin R1

=

C R2 s

+

1 R2

1 R

1 R2

1 R2

1 R

1 R1 Vin

+ Cs

0

C R1 sVin 1 RR1 Vin

Plug in V1 , V2 and V3 to the fourth equation.

=

V3 V2 V1 Vin + + + Ra Rb Rc Rd 1 1 1 1 + V2 + V1 + Vin Ra Rb Rc Rd

=

C 1 1 1 RR1 R1 s V + Vin + Vin in 1 1 C C 2 2 2 2 Rc C s + R s + R 2 Rd C s + R2 s + R2 2 # C 1 1 1 1 RR1 R1 s + + Vin Rb C 2 s2 + RC s + R12 Rc C 2 s2 + RC s + R12 Rd 2 2

1 1 + Ra Rb

= "

1 + Ra Vout R

= Finally,

Vout Vin

=

R

"

=

R

=

R C2

1 1 + Ra Rb 1 Ra

+

1 Rb

1 RR1 C 2 s2 + RC2 s

1 RR1

1 C Rc R1 s

C 2 s2 + C2 2 Rd s

+

1 C Rd R2

1 C Rc R1

s2 +

+

1 R2

1 Rd

+

C R2 s

+

s+

1 R2 C s

+

C 1 R1 s + Rc C 2 s2 + RC s + 2

C 2 s2 +

C R2 s

+

1 R2

1 RR1

+

1 1 Rd R2

1 R2 1 Rb 1 (RC)2

1 Ra

1 R2

1 + Rd

#

2029 (a) If Ra = R; and Rb = Rc = Rd = 1; Vout Vin

C2 2 Rd s

=

R C2

=

R C 2 s2 +

=

+

1 C Rd R2

1 C Rc R1

s2 + 1 1 R RR1 1 1 R2 C s + (RC)2

=

s+

1 R2 C s

s2 +

1 Rb

1 Ra

1 RR1

+

1 1 Rd R2

1 (RC)2 1 RR1 C 2 1 1 R2 C s + (RC)2

+

R R1 2

(RC) s2 +

R2 C R2 s

+1

So, R R1 2

!n

= =

= A

(RC)

=

2

=

!n

1 ! 2n R2 C R2

1 RC ! n R2 C 1 R2 C R = = 2 R2 2RC R2 2R2

(b) Step response using MatLab % Problem 2.16 A = 1; wn = 1; z = [ 0.1 0.5 1.0 ]; hold on for i = 1:3 num = [ A ]; den = [ 1/wn^2 2*z(i)/wn 1 ] step( num, den ) end hold o¤ 17. Find the equations and transfer function for the biquad circuit of Fig. 2.49 if Ra = R; Rd = R1 and Rb = Rc = 1: Solution:

2030

CHAPTER 2. DYNAMIC MODELS

Vout Vin

C2 2 Rd s

+

1 C Rd R2

1 C Rc R1

=

R C2

=

R C2

=

s2 + R21C s R 1 R1 s2 + R21C s + (RC) 2

s2 + C2 2 R1 s

+

1 C R1 R2

s2 +

1 R2 C s 1 R

s+ 1 R2 C s

s+

+

+

1 RR1

1 (RC)2

1 Rb

1 Ra

1 (RC)2

+

1 1 R1 R2

1 RR1

+

1 1 Rd R2

2031

Problems and Solutions for Section 2.3 18. The torque constant of a motor is the ratio of torque to current and is often given in ounce-inches per ampere. (ounce-inches have dimension force-distance where an ounce is 1=16 of a pound.) The electric constant of a motor is the ratio of back emf to speed and is often given in volts per 1000 rpm. In consistent units the two constants are the same for a given motor. (a) Show that the units ounce-inches per ampere are proportional to volts per 1000 rpm by reducing both to MKS (SI) units. (b) A certain motor has a back emf of 25 V at 1000 rpm. What is its torque constant in ounce-inches per ampere? (c) What is the torque constant of the motor of part (b) in newton-meters per ampere? Solution: Before going into the problem, let’s review the units. Some remarks on non SI units. – Ounce 1oz = 2:835

10

2

kg

Originally ounce is a unit of mass, but like pounds, it is commonly used as a unit of force. If we translate it as force,

1oz(f) = 2:835

2

10

kgf = 2:835

10

2

2

m

9:81 N = 0:2778 N

– Inch 1 in = 2:540

10

– RPM (Revolution per Minute) 1 RPM =

2

rad = rad/ s 60 s 30

Relation between SI units – Voltage and Current

V olts Current(amps) V olts

= P ower = Energy(joules)= sec Joules= sec N ewton meters= sec = = amps amps

2032

CHAPTER 2. DYNAMIC MODELS (a) Relation between torque constant and electric constant. Torque constant: 1 ounce 1 inch 0:2778 N = 1 Ampere

2:540 1A

10

2

m

= 7:056 10

3

N m= A

Electric constant: 1V 1 J=(A sec) = = 9:549 1000 RPM 1000 30 rad/ s

10

3

N m= A

So,

1 oz in= A

= =

7:056 10 3 V=1000 RPM 9:549 10 3 (0:739) V=1000 RPM

(b) 25 V=1000 RPM = 25

1 oz in= A = 33:872 oz in= A 0:739

(c) 25 V=1000 RPM = 25

9:549

10

3

N m= A = 0:239 N m= A

19. The electromechanical system shown in Fig. 2.50 represents a simpli…ed model of a capacitor microphone. The system consists in part of a parallel plate capacitor connected into an electric circuit. Capacitor plate a is rigidly fastened to the microphone frame. Sound waves pass through the mouthpiece and exert a force fs (t) on plate b, which has mass M and is connected to the frame by a set of springs and dampers. The capacitance C is a function of the distance x between the plates, as follows: C(x) =

"A ; x

where " = dielectric constant of the material between the plates; A = surface area of the plates: The charge q and the voltage e across the plates are related by q = C(x)e: The electric …eld in turn produces the following force fe on the movable plate that opposes its motion: fe =

q2 2"A

2033 (a) Write di¤erential equations that describe the operation of this system. (It is acceptable to leave in nonlinear form.) (b) Can one get a linear model? (c) What is the output of the system? Figure 2.48: Simpli…ed model for capacitor microphone

Solution: (a) The free body diagram of the capacitor plate b

− Kx − Bx&

M

− sgn(x& )f e

f s (t )

x Free body diagram So the equation of motion for the plate is Mx • + B x_ + Kx + fe sgn (x) _ = fs (t) : The equation of motion for the circuit is d i+e dt where e is the voltage across the capacitor, Z 1 e= i(t)dt C v = iR + L

2034

CHAPTER 2. DYNAMIC MODELS and where C = "A=x; a variable. Because i = can rewrite the circuit equation as v = Rq_ + L• q+

d dt q

and e = q=C; we

qx "A

In summary, we have these two, couptled, non-linear di¤erential equation. q2 2"A qx Rq_ + L• q+ "A

Mx • + bx_ + kx + sgn (x) _

= fs (t) = v

(b) The sgn function, q 2 , and qx; terms make it impossible to determine a useful linearized version. (c) The signal representing the voice input is the current, i, or q: _ 20. A very typical problem of electromechanical position control is an electric motor driving a load that has one dominant vibration mode. The problem arises in computer-disk-head control, reel-to-reel tape drives, and many other applications. A schematic diagram is sketched in Fig. 2.51. The motor has an electrical constant Ke , a torque constant Kt , an armature inductance La , and a resistance Ra . The rotor has an inertia J1 and a viscous friction B. The load has an inertia J2 . The two inertias are connected by a shaft with a spring constant k and an equivalent viscous damping b. Write the equations of motion.

Figure 2.49: Motor with a ‡exible load

(a)

Solution: (a) Rotor: J1 •1 =

B _1

b _1

_2

k(

1

2)

+ Tm

2035 Load:

J2 •2 =

b _2

_1

k(

2

1)

Circuit:

va

Ke _ 1 = La

d ia + Ra ia dt

Relation between the output torque and the armature current:

Tm = Kt ia

2036

CHAPTER 2. DYNAMIC MODELS

Problems and Solutions for Section 2.4 21. A precision-table leveling scheme shown in Fig. 2.52 relies on thermal expansion of actuators under two corners to level the table by raising or lowering their respective corners. The parameters are: Tact = actuator temperature; Tamb = ambient air temperature; Rf = heat ow coe cient between the actuator and the air; C = thermal capacity of the actuator; R = resistance of the heater: Assume that (1) the actuator acts as a pure electric resistance, (2) the heat ‡ow into the actuator is proportional to the electric power input, and (3) the motion d is proportional to the di¤erence between Tact and Tamb due to thermal expansion. Find the di¤erential equations relating the height of the actuator d versus the applied voltage vi . Figure 2.50: (a) Precision table kept level by actuators; (b) side view of one actuator

Solution: Electric power in is proportional to the heat ‡ow in 2

v Q_ in = Kq i R and the heat ‡ow out is from heat transfer to the ambient air 1 Q_ out = (Tact Tamb ) : Rf

2037 The temperature is governed by the di¤erence in heat ‡ows T_act

= =

1 _ Qin Q_ out C 1 v2 1 Kq i (Tact C R Rf

Tamb )

and the actuator displacement is d = K (Tact

Tamb ) :

where Tamb is a given function of time, most likely a constant for a table inside a room. The system input is vi and the system output is d: 22. An air conditioner supplies cold air at the same temperature to each room on the fourth ‡oor of the high-rise building shown in Fig. 2.53(a). The ‡oor plan is shown in Fig. 2.53(b). The cold air ‡ow produces an equal amount of heat ‡ow q out of each room. Write a set of di¤erential equations governing the temperature in each room, where To = temperature outside the building; Ro = resistance to heat ow through the outer walls; Ri = resistance to heat ow through the inner walls: Assume that (1) all rooms are perfect squares, (2) there is no heat ‡ow through the ‡oors or ceilings, and (3) the temperature in each room is uniform throughout the room. Take advantage of symmetry to reduce the number of di¤erential equations to three. Solution: We can classify 9 rooms to 3 types by the number of outer walls they have. Type 1 Type 2 Type 1

Type 2 Type 3 Type 2

Type 1 Type 2 Type 1

We can expect the hotest rooms on the outside and the corners hotest of all, but solving the equations would con…rm this intuitive result. That is, To > T1 > T2 > T3 and, with a same cold air ‡ow into every room, the ones with some sun load will be hotest. Let’s rede…nce the resistances

Ro = resistance to heat ow through one unit of outer wall Ri = resistance to heat ow through one unit of inner wall

2038

CHAPTER 2. DYNAMIC MODELS

Figure 2.51: Building air conditioning: (a) high-rise building, (b) ‡oor plan of the fourth ‡oor

Room type 1:

T_1

= =

qout

=

qin

=

2 (T1 Ri 2 (To Ro

1 (qin qout ) C 1 2 (To T1 ) C Ro

T2 ) + q T1 )

2 (T1 Ri

T2 )

q

Room type 2:

1 T_2 = C

qin

=

qout

=

1 (To Ro

1 (To Ro 1 (T2 Ri T2 ) +

T2 ) +

2 (T1 Ri

T2 )

T3 ) + q

2 (T1 Ri

T2 )

1 (T2 Ri

T3 )

q

2039 Room type 3:

qout

4 (T2 Ri = q

1 T_3 = C

4 (T2 Ri

qin

=

T3 )

T3 )

q

23. For the two-tank ‡uid-‡ow system shown in Fig. 2.54, …nd the di¤erential equations relating the ‡ow into the …rst tank to the ‡ow out of the second tank. Figure 2.52: Two-tank ‡uid-‡ow system for Problem 23

Solution: This is a variation on the problem solved in Example 2.18 and the de…nitions of terms is taken from that. From the relation between the height of the water and mass ‡ow rate, the continuity equations are m _1 m _2

= =

A1 h_ 1 = win w A2 h_ 2 = w wout

Also from the relation between the pressure and outgoing mass ‡ow rate, w

=

wout

=

1 1 ( gh1 ) 2 R1 1 1 ( gh2 ) 2 R2

2040

CHAPTER 2. DYNAMIC MODELS Finally, h_ 1

=

h_ 2

=

1 1 1 ( gh1 ) 2 + win A1 R1 A1 1 1 1 1 ( gh1 ) 2 ( gh2 ) 2 : A2 R1 A2 R2

24. A laboratory experiment in the ‡ow of water through two tanks is sketched in Fig. 2.55. Assume that Eq. (2.74) describes ‡ow through the equal-sized holes at points A, B, or C. (a) With holes at A and C but none at B, write the equations of motion for this system in terms of h1 and h2 . Assume that h3 = 20 cm, h1 > 20 cm; and h2 < 20 cm. When h2 = 10 cm, the out‡ow is 200 g/min. (b) At h1 = 30 cm and h2 = 10 cm, compute a linearized model and the transfer function from pump ‡ow (in cubic centimeters per minute) to h2 . (c) Repeat parts (a) and (b) assuming hole A is closed and hole B is open.

Figure 2.53: Two-tank ‡uid-‡ow system for Problem 24

Solution:

2041 (a) Following the solution of Example 2.18, and assuming the area of both tanks is A; the values given for the heights ensure that the water will ‡ow according to 1 1 WA = [ g (h1 h3 )] 2 R 1 1 WC = [ gh2 ] 2 R WA WC = Ah_ 2 Win WA = Ah_ 1 From the out‡ow information given, we can compute the ori…ce resistance, R; noting that for water, = 1 gram/cc and g = 981 cm/sec2 ' 1000 cm/sec2 : 1p 1p WC = 200 g= mn = gh2 = g 10 cm R p R p 1 g= cm3 1000 cm= s2 10 cm g 10 cm R = = 200 g= mn 200 g=60 s s 1 1 g cm2 s2 100 = 60 = 30 g 2 cm 2 3 2 2 200 cm s g (b) The nonlinear equations from above are h_ 1

=

h_ 2

=

1 1 p g (h1 h3 ) + Win AR A 1 p 1 p g (h1 h3 ) gh2 AR AR

The square root functions need to be linearized about the nominal heights. In general the square root function can be linearized as below p x0 + x = =

s p

x0 1 +

x x0

x0 1 +

1 x 2 x0

So let’s assume that h1 = h10 + h1 and h2 = h20 + h2 where h10 = 30 cm, h20 = 10 cm, and h3 = 20 cm. And for round numbers, let’s assume the area of each tank A = 100 cm2 : The equations above then reduce to p 1 1 h_ 1 = (1)(1000) (30 + h1 20) + Win (1)(100)(30) (1)(100) p p 1 1 h_ 2 = (1)(1000) (30 + h1 20) (1)(1000)(10 + h2 ) (1)(100)(30) (1)(100)(30)

2042

CHAPTER 2. DYNAMIC MODELS which, with the square root approximations, is equivalent to, h_ 1

=

h_ 2

=

1 1 1 (1 + h1 ) + Win (30) 20 (100) 1 1 1 1 (1 + h1 ) (1 + h2 ) (30) 20 (30) 20

The nominal in‡ow Wnom = 10 3 cc/sec is required in order for the system to be in equilibrium, as can be seen from the …rst equation. So we will de…ne the total in‡ow to be Win = Wnom + W; so the equations become h_ 1

=

h_ 2

=

1 1 1 1 (1 + h1 ) + Wnom + W (30) 20 (100) (100) 1 1 1 1 (1 + h1 ) (1 + h2 ) (30) 20 (30) 20

or, with the nominal in‡ow included, the equations reduce to h_ 1

=

h_ 2

=

1 1 h1 + W 600 100 1 1 h1 h2 600 600

Taking the Laplace transform of these two equations, and solving for the desired transfer function (in cc/sec) yields H2 (s) 1 0:01 = : W (s) 600 (s + 1=600)2 which becomes, with the in‡ow in grams/min, H2 (s) 1 (0:01)(60) 0:001 = =: W (s) 600 (s + 1=600)2 (s + 1=600)2 (c) With hole B open and hole A closed, the relevant relations are Win

WB

h_ 1

=

h_ 2

=

WB

=

WB

=

WC

=

WC

=

Ah_ 1 1p g(h1 R Ah_ 2 1p gh2 R

h2 )

1 p 1 g(h1 h2 ) + Win AR A 1 p 1 p g(h1 h2 ) gh2 AR AR

2043 With the same de…nitions for the perturbed quantities as for part (b), we obtain h_ 1

=

h_ 2

=

p 1 1 (1)(1000)(30 + h1 10 h2 ) + Win (1)(100)(30) (1)(100) p 1 (1)(1000)(30 + h1 10 h2 ) (1)(100)(30) p 1 (1)(1000)(10 + h2 ) (1)(100)(30)

which, with the linearization carried out, reduces to p 1 1 1 2 _h1 = (1 + h1 h2 ) + Win 30 40 40 100 p 1 1 1 1 2 h_ 2 = (1 + h1 h2 ) (1 + h2 ) 30 40 40 30 20 and with the nominal ‡ow rate of Win = h_ 1

=

h_ 2

=

p 10 2 3

removed

p

1 2 ( h1 h2 ) + W 1200 100 p p p 1 2 2 2 1 h1 + ( ) h2 + 1200 1200 600 30

However, unlike part (b), holding the nominal ‡ow rate maintains h1 at equilibrium, but h2 will not stay at equilibrium. Instead, there will be a constant term increasing h2 : Thus the standard transfer function will not result. 25. The equations for heating a house are given by Eqs. (2.62) and (2.63) and, in a particular case can be written with time in hours as C

dTh = Ku dt

Th

To R

where (a) C is the Thermal capacity of the house, BT U=o F (b) Th is the temperature in the house,

o

F

(c) To is the temperature outside the house,

o

F

(d) K is the heat rating of the furnace, = 90; 000 BT U=hour (e) R is the thermal resistance, o F per BT U=hour (f) u is the furnace switch, =1 if the furnace is on and =0 if the furnace is o¤.

2044

CHAPTER 2. DYNAMIC MODELS It is measured that, with the outside temperature at 32 o F and the house at 60 o F , the furnace raises the temperature 2 o F in 6 minutes (0.1 hour). With the furnace o¤, the house temperature falls 2 o F in 40 minutes. What are the values of C and R for the house? Solution: For the …rst case, the furnace is on which means u = 1.

C

dTh dt T_h

= K =

K C

1 (Th To ) R 1 (Th To ) RC

and with the furnace o¤, 1 (Th To ) RC In both cases, it is a …rst order system and thus the solutions involve exponentials in time. The approximate answer can be obtained by simply looking at the slope of the exponential at the outset. This will be fairly accurate because the temperature is only changing by 2 degrees and this represents a small fraction of the 30 degree temperature di¤erence. Let’s solve the equation for the furnace o¤ …rst T_h =

Th 1 = (Th To ) t RC plugging in the numbers available, the temperature falls 2 degrees in 2/3 hr, we have 2 1 = (60 32) 2=3 RC which means that RC = 28=3 For the second case, the furnace is turned on which means Th K 1 = (Th t C RC and plugging in the numbers yields 2 90; 000 = 0:1 C

1 (60 28=3

To )

32)

and we have C

=

R

=

90; 000 = 3910 23 RC 28=3 = = 0:00240 C 3910

Chapter 3

Dynamic Response Problems and Solutions for Section 3.1: Review of Laplace Transforms 1. Show that, in a partial-fraction expansion, complex conjugate poles have coe¢ cients that are also complex conjugates. (The result of this relationship is that whenever complex conjugate pairs of poles are present, only one of the coe¢ cients needs to be computed.) Solution: Consider the second-order system with poles at H(s) =

(s +

1 + j ) (s +

j , j )

:

Perform Partial Fraction Expansion: H(s) C1 C2

C1 C2 + : s+ +j s+ j 1 1 = js= j; j = s+ j 2 1 1 = js= +j = j; s+ +j 2 ) C1 = C2 : =

2. Find the Laplace transform of the following time functions: (a) f (t) = 1 + 2t (b) f (t) = 3 + 7t + t2 + (t) (c) f (t) = e

t

+ 2e

2t

+ te

3t

(d) f (t) = (t + 1)2 3001

3002

CHAPTER 3. DYNAMIC RESPONSE (e) f (t) = sinh t Solution: (a) f (t) = 1 + 2t: Lff (t)g = Lf1(t)g + Lf2tg; 1 2 = + 2; s s s+2 : = s2 We can verify the answer using Matlab: >> laplace(1+2*t) ans = (2+s)/s^2 (b) f (t) = 3 + 7t + t2 + (t); Lff (t)g = Lf3g + Lf7tg + Lft2 g + Lf (t)g; 3 7 2! = + 2 + 3 + 1; s s s s3 + 3s2 + 7s + 2 : = s3 We can verify the answer using Matlab: >> laplace(3+7*t+t^2+dirac(t)) ans = 1+3/s+7/s^2+2/s^3 (c) f (t) = e t + 2e 2t + te 3t ; Lff (t)g = Lfe t g + Lf2e 2t g + Lfte 1 2 1 = + + : s + 1 s + 2 (s + 3)2 We can verify the answer using Matlab: >> laplace(exp(-t)+2*exp(-2*t)+t*exp(-3*t)) ans = 1/(1+s)+2/(2+s)+1/(s+3)^2

3t

g;

3003 (d) = (t + 1)2 ; = t2 + 2t + 1: Lff (t)g = Lft2 g + Lf2tg + Lf1g; 2! 2 1 = + 2+ ; 3 s s s s2 + 2s + 2 : = s3 f (t)

We can verify the answer using Matlab: >> laplace((t+1)^2) ans = (2+2*s+s^2)/s^3 (e) Using the trigonometric identity, f (t)

=

sinh t; et e t ; 2 t e e t L L ; 2 2 1 1 1 1 2 s 1 2 s+1 1 : s2 1

= Lff (t)g = = =

;

We can verify the answer using Matlab: >> laplace(sinh(t)) ans = 1/(s^2-1) Remark: A useful reference for this problem and the next several problems is: K. R. Coombes, B. R. Hunt, R. L. Lipsman, J. E. Osborn, G. J. Stuck, Di¤ erential Equations with Matlab, Wiley, 1998. 3. Find the Laplace transform of the following time functions: (a) f (t) = 3 cos 6t (b) f (t) = sin 2t + 2 cos 2t + e (c) f (t) = t2 + e Solution:

2t

sin 3t

t

sin 2t

3004

CHAPTER 3. DYNAMIC RESPONSE (a) f (t) = 3 cos 6t Lff (t)g = Lf3 cos 6tg s = 3 2 : s + 36 We can verify the answer using Matlab: >> laplace(3*cos(6*t)) ans = 3*s/(s^2+36) (b) f (t)

= sin 2t + 2 cos 2t + e t sin 2t = Lff (t)g = Lfsin 2tg + Lf2 cos 2tg + Lfe 2s 2 2 + + : = s2 + 4 s2 + 4 (s + 1)2 + 4

t

We can verify the answer using Matlab: >> laplace(sin(2*t)+2*cos(2*t)+exp(-t)*sin(2*t)) ans = 2*(9+7*s+4*s^2+s^3)/(s^2+4)/(5+2*s+s^2) (c) f (t)

= t2 + e 2t sin 3t; = Lff (t)g = Lft2 g + Lfe 2! 3 + ; = 3 s (s + 2)2 + 9 2 3 = + : s3 (s + 2)2 + 9

2t

sin 3tg;

We can verify the answer using Matlab: >> laplace(t^2+exp(-2*t)*sin(3*t)) ans = 2/s^3+3/(s^2+4*s+13) 4. Find the Laplace transform of the following time functions: (a) f (t) = t sin t (b) f (t) = t cos 3t (c) f (t) = te

t

+ 2t cos t

(d) f (t) = t sin 3t

2t cos t

(e) f (t) = 1(t) + 2t cos 2t

sin 2tg;

3005 Solution: (a) f (t) = t sin t Lff (t)g = Lft sin tg Use multiplication by time Laplace transform property (Table A.1, entry #11), Lftg(t)g = Let g(t)

=

Lft sin tg

= = =

d G(s): ds sin t

and use

d 1 2 ds s + 12 2s ; (s2 + 1)2 2s : s4 + 2s2 + 1

Lfsin atg =

a : s2 + a2

;

We can verify the answer using Matlab: >> laplace(t*sin(t)) ans = 2*s/(s^2+1)^2 (b) f (t) = t cos 3t Use multiplication by time Laplace transform property (Table A.1, entry #11), Lftg(t)g = Let g(t)

=

Lft cos 3tg = = =

d G(s): ds cos 3t

and use

Lfcos atg =

d s ; 2 ds s + 9 [(s2 + 9) (2s)s] ; (s2 + 9)2 s2 9 : s4 + 18s2 + 81

We can verify the answer using Matlab: >> laplace(t*cos(3*t)) ans = (s^2-9)/(s^2+9)^2

s : s2 + a2

3006

CHAPTER 3. DYNAMIC RESPONSE (c) f (t) = te

t

+ 2t cos t

Use the following Laplace transforms and properties (Table A.1, entries 4,11, and 3), Lfte

at

g =

Lftg(t)g = Lfcos atg = Lff (t)g = = = =

1 ; (s + a)2 d G(s); ds s ; s2 + a2 Lfte t g + 2Lft cos tg; s d 1 +2 ; (s + 1)2 ds s2 + 1 (s2 + 1) (2s)s 1 2 ; 2 (s + 1) (s2 + 1)2 2(s2 1) 1 + 2 : 2 (s + 1) (s + 1)2

We can verify the answer using Matlab: >> laplace(t*exp(-t)+2*t*cos(t)) ans = 1/(1+s)^2+2*(s^2-1)/(s^2+1)^2 (d) f (t) = t sin 3t

2t cos t:

Use the following Laplace transforms and properties (Table A.1, entries 11, 3), Lftg(t)g = Lfsin atg = Lfcos atg = Lff (t)g = = = =

d G(s); ds a ; s2 + a2 s ; 2 s + a2 Lft sin 3tg 2Lft cos tg; d 3 d s 2 2 2 ds s + 9 ds s + 1 2 (2s 3) [(s + 1) (2s)s] +2 ; (s2 + 9)2 (s2 + 1)2 6s 2(s2 1) : 2 2 (s + 9) (s2 + 1)2

We can verify the answer using Matlab:

;

3007 >> laplace(t*sin(3*t)-2*t*cos(t)) ans = 6*s/(s^2+9)^2-2*(s^2-1)/(s^2+1)^2 (e) f (t)

=

Lf1(t)g = Lftg(t)g = Lfcos atg = Lff (t)g = = = =

1(t) + 2t cos 2t; 1 ; s d G(s); ds s ; s2 + a2 Lf1(t)g + 2Lft cos 2tg; 1 d s +2 ; 2 s ds s + 4 (s2 + 4) (2s)s 1 ; 2 s (s2 + 4)2 ( s2 + 4) 1 2 2 : s (s + 4)2

We can verify the answer using Matlab: >> laplace(1+2*t*cos(2*t)) ans = 1/s+2*(s^2-4)/(s^2+4)^2 5. Find the Laplace transform of the following time functions (* denotes convolution): (a) f (t) = sin t sin 3t (b) f (t) = sin2 t + 3 cos2 t (c) f (t) = (sin t)=t (d) f (t) = sin t sin t Rt (e) f (t) = 0 cos(t ) sin d Solution: (a) f (t) = sin t sin 3t:

3008

CHAPTER 3. DYNAMIC RESPONSE Use the trigonometric relation, sin t sin t

= =

f (t)

= =

Lff (t)g = = =

1 1 cos(j jt) cos(j + jt); 2 2 1 and = 3: 1 1 cos(j1 3jt) cos(j1 + 3jt); 2 2 1 1 cos 2t sin 4t: 2 2 1 1 Lfcos 2tg Lfcos 4tg; 2 2 1 s s : 2 s2 + 4 s2 + 16 6s : (s2 + 4)(s2 + 16)

We can verify the answer using Matlab: >> laplace(sin(t)*sin(3*t)) ans = 6*s/(s^2+16)/(s^2+4) (b) f (t) = sin2 t + 3 cos2 t: Use the trigonometric formulas, sin2 t

=

cos2 t

=

f (t)

=

= Lff (t)g = = =

1

cos 2t ; 2 1 + cos 2t ; 2 1 + cos 2t 1 cos 2t +3 2 2 2 + cos 2t: Lf2g + Lfcos 2tg 2 s + 2 ; s s +4 3s2 + 8 : s(s2 + 4)

We can verify the answer using Matlab: >> laplace(sin(t)^2+3*cos(t)^2) ans = (8+3*s^2)/s/(s^2+4)

;

3009 (c) We …rst show the result that division by time is equivalent to integration in the frequency domain. This can be done as follows, Z 1 e st f (t)dt; F (s) = Z0 1 Z 1 Z 1 e st f (t)dt ds; F (s)ds = 0

s

s

Interchanging the order of integration, Z 1 Z 1 Z 1 e st ds f (t)dt; F (s)ds = s s 0 Z 1 Z 1 1 1 st F (s)ds = e f (t)dt; t s 0 s Z 1 f (t) st e dt: = t 0 Using this result then, Lfsin tg = L

sin t t

= =

s2

Z

1

1 d ; +1 s tan 1 (1) tan

= =

1 ; +1 2

1

tan

2 tan

1

1 s

1

(s);

(s); :

where a table of integrals was used and the last simpli…cation follows from the related trigonometric identity. (d) f (t) = sin t sin t: Use the convolution Laplace transform property (Table A.1, entry 7), Lfsin t sin tg = =

1 1 2 +1 s +1 1 : s4 + 2s2 + 1 s2

;

(e) f (t)

=

Z

t

cos(t ) sin d : Z t Lff (t)g = L cos(t ) sin d 0

0

= Lfcos(t) sin(t)g:

3010

CHAPTER 3. DYNAMIC RESPONSE This is just the de…nition of the convolution theorem, s 1 ; s2 + 1 s2 + 1 s : 4 s + 2s2 + 1

Lff (t)g = =

6. Given that the Laplace transform of f(t) is F(s), …nd the Laplace transform of the following: (a) g(t) = f (t) cos t RtRt (b) g(t) = 0 0 1 f ( )d dt1

Solution:

(a) First write cos t in terms of the related Euler identity (Eq. B.33), g(t) = f (t) cos t = f (t)

ejt + e 2

jt

=

1 1 f (t)ejt + f (t)e 2 2

jt

:

Then using entry 4 of Table A.1 we have, G(s) =

1 F (s 2

1 1 j) + F (s + j) = [F (s 2 2

(b) Let us de…ne ef (t1 ) =

then

g(t) =

Z

j) + F (s + j)] :

t1

f ( )d ;

0

Z

0

t

fe(t1 )dt1 ;

and from entry 6 of Table A.1 we have

Lffe(t)g = Fe(s) =

1 F (s) s

and using the same result again, we have G(s) =

1e 1 1 1 F (s) = F (s) = 2 F (s): s s s s

7. Find the time function corresponding to each of the following Laplace transforms using partial fraction expansions: (a) F (s) =

2 s(s+2)

(b) F (s) =

10 s(s+1)(s+10)

(c) F (s) =

3s+2 s2 +4s+20

3011 (d) F (s) =

3s2 +9s+12 (s+2)(s2 +5s+11)

(e) F (s) =

1 s2 +4

(f) F (s) =

2(s+2) (s+1)(s2 +4)

(g) F (s) =

s+1 s2

(h) F (s) =

1 s6

(i) F (s) =

4 s4 +4

(j) F (s) =

e s s2

Solution:

(a) Perform partial fraction expansion,

F (s)

= =

L

1

C1

=

C2

=

F (s)

=

fF (s)g = f (t)

=

2 ; s(s + 2) C1 C2 + : s s+2 2 js=0 = 1; s+2 2 js= 2 = 1; s 1 1 : s s+2 1 L L 1 s 1(t)

e

2t

1(t):

We can verify the answer using Matlab: >> ilaplace(2/(s*(s+2))) ans = 1-exp(-2*t)

1

1 s+2

;

3012

CHAPTER 3. DYNAMIC RESPONSE (b) Perform partial fraction expansion, F (s)

= =

C1

=

C2

=

C3

=

F (s)

=

f (t)

=

10 ; s(s + 1)(s + 10) C1 C2 C3 + + : s s + 1 s + 10 10 js=0 = 1; (s + 1)(s + 10) 10 10 js= 1 = ; s(s + 10) 9 1 10 js= 10 = ; s(s + 1) 9 10 1 1 9 + 9 ; s s + 1 s + 10 1 10 t e 1(t) + e L 1 fF (s)g = 1(t) 9 9

10t

1(t):

We can verify the answer using Matlab: >> ilaplace(10/(s*(s+1)*(s+10))) ans = -10/9*exp(-t)+1+1/9*exp(-10*t) (c) Re-write and carry out partial fraction expansion, F (s)

f (t)

3s + 2 ; s2 + 4s + 20 4 (s + 2) 3 ; = 3 (s + 2)2 + 42 4 3(s + 2) ; = 2 2 (s + 2) + 4 (s + 2)2 + 42 =

= L

1

fF (s)g = (3e

2t

cos 4t

e

2t

sin 4t)1(t):

We can verify the answer using Matlab: >> ilaplace((3*s+2)/(s^2+4*s+20)) ans = exp(-2*t)*(3*cos(4*t)-sin(4*t)) (d) Perform partial fraction expansion, F (s)

= =

C1

=

3s2 + 9s + 12 (s + 2)(s2 + 5s + 11) C1 C2 s + C3 + s + 2 s2 + 5s + 11 3(s2 + 3s + 4) 6 js= 2 = : (s2 + 5s + 11) 5

3013 Equate numerators: 6 C2 s + C3 3s2 + 9s + 12 5 = ; + 2 (s + 2) (s + 5s + 11) (s + 2)(s2 + 5s + 11) 6 66 (C2 + )s2 + (6 + C3 + 2C2 )s + (2C3 + ) = 3s2 + 9s + 12: 5 5 Equate like powers of s to …nd C2 and C3 : 6 5 66 2C3 + 5 C2 +

F (s)

9 ; 5

=

3 ) C2 =

=

12 ) C3 =

=

3 ; 5 9 3 6 s 5 5 5 + ; (s + 2) (s2 + 5s + 11)

p 5 19 p s+ 19 9 17 2 2 = p !2 : 2 2 5 57 19 5 5 19 + s+ s+ + 2 4 2 2 0 1 p p p 5 5 t t 6 9 19 153 19 19 f (t) = L 1 fF (s)g = @ e 2t + e 2 cos t e 2 sin tA 1(t): 5 5 2 285 2 6 9 5 + (s + 2) 5

We can verify the answer using Matlab: >> ilaplace((3*s^2+9*s+12)/((s+2)*(s^2+5*s+11))) ans = 6/5*exp(-2*t)+3/95*exp(-5/2*t)*(57*cos(1/2*19^(1/2)*t)-17*19^(1/2)*sin(1/2*19^(1/2)*t)) (e) Re-write and use entry #17 of Table A.2, F (s)

= =

f (t)

=

1 : s2 + 4 1 2 : 2 2 (s + 22 ) 1 sin 2t: 2

We can verify the answer using Matlab: >> ilaplace(1/(s^2+4)) ans = 1/2*sin(2*t)

3014

CHAPTER 3. DYNAMIC RESPONSE (f) F (s)

= =

C1

=

2(s + 2) : (s + 1)(s2 + 4) C1 C2 s + C3 + 2 : (s + 1) (s + 4) 2(s + 2) 2 js= 1 = : 2 (s + 4) 5

Equate numerators and like powers of s terms: 8 + C3 5 8 + C3 5

2 + C2 s2 + (C2 + C3 )s + 5

F (s)

=

= f (t)

=

2 5 (s + 1) 2 5 (s + 1) 2 t e 5

=

2s + 4;

=

4

C2 + C3

=

2

2 + C2 5

=

0:

12 2 s+ 5 ; + 52 (s + 4) 2 s 6 2 + 2 5 2 + : 2 (s + 2 ) 5 (s + 22 ) 2 6 cos 2t + sin 2t: 5 5

We can verify the answer using Matlab: >> ilaplace(2*(s+2)/((s+1)*(s^2+4))) ans = -4/5*cos(t)^2+12/5*sin(t)*cos(t)+2/5+2/5*exp(-t) (g) Perform partial fraction expansion, F (s)

= =

f (t)

12 ; 5 2 ) C2 = ; 5 ) C3 =

=

s+1 ; s2 1 1 + 2: s s (1 + t)1(t):

We can verify the answer using Matlab: >> ilaplace((s+1)/(s^2)) ans = t+1

3015 (h) Use entry #6 of Table A.2, 1 ; s6

F (s)

=

f (t)

= L

1

1 s6

=

t5 t5 = : 5! 120

We can verify the answer using Matlab: >> ilaplace(1/s^6) ans = 1/120*t^5 (i) Re-write as, F (s)

= = =

4 ; s4 + 4 1 1 s+1 2s + 1 + 2 2 ; 2 s + 2s + 2 s 2s + 2 (s + 1) 12 s (s 1) 12 s : (s + 1)2 + 1 (s 1)2 + 1

Use Table A.2 entry #19 and Table A.1 entry #5,

f (t)

= L

1

fF (s)g = e

t

cos(t)

1 d e 2 dt

t

1 d t fe sin(t)g; 2 dt 1 = e t cos(t) f e t sin(t) + cos(t)e t g 2 1 et cos(t) + fet sin(t) + cos(t)et g; 2 e t + et e t + et = cos(t) + sin(t) 2 2 f (t) = cos(t) sinh(t) + sin(t) cosh(t): We can verify the answer using Matlab: >> ilaplace(4/(s^4+4)) ans = cosh(t)*sin(t)-sinh(t)*cos(t) (j) Using entry #2 of Table A.1, F (s) f (t)

e s : s2 1 = L fF (s)g = (t

et cos(t);

sin(t)

=

1)1(t

1):

;

3016

CHAPTER 3. DYNAMIC RESPONSE We can verify the answer using Matlab: >> ilaplace(exp(-s)/(s^2)) ans = heaviside(t-1)*(t-1)

8. Find the time function corresponding to each of the following Laplace transforms: (a) F (s) =

1 s(s+2)2

(b) F (s) =

2s2 +s+1 s3 1

(c) F (s) =

2(s2 +s+1) s(s+1)2

(d) F (s) =

s3 +2s+4 s4 16

(e) F (s) =

2(s+2)(s+5)2 (s+1)(s2 +4)2

(f) F (s) =

(s2 1) (s2 +1)2

(g) F (s) = tan

1 1 (s)

Solution: (a) Perform partial fraction expansion, F (s)

= =

C1

=

C3

=

C2

= = = =

F (s)

=

f (t)

=

1 ; s(s + 2)2 C2 C3 C1 + + : s (s + 2) (s + 2)2 1 1 sF (s)js=0 = js=0 = ; 2 (s + 2) 4 1 1 2 (s + 2) F (s)js= 2 = js= 2 = ; s 2 d [(s + 2)2 F (s)]s= 2 ; ds d [s 1 ]s= 2 ; ds 1 js= 2 ; s2 1 ; 4 1 1 1 4 + 4 + 2 : s (s + 2) (s + 2)2 1 1 2t 1 L 1 fF (s)g = e te 2t 1(t): 4 4 2

3017 We can verify the answer using Matlab: >> ilaplace(1/(s*(s+2)^2)) ans = 1/4-1/4*exp(-2*t)*(1+2*t) (b) Perform partial fraction expansion, F (s)

2s2 + s + 1 ; s3 1 2s2 + s + 1 ; (s 1)(s2 + s + 1) C1 C2 s + C3 + : s 1 s2 + s + 1 4 2s2 + s + 1 js=1 = : (s 1)F (s)js=1 = 2 s +s+1 3

= = =

C1

=

Equate numerators and match the coe¢ cients of like powers of s: 4 3

C2 s + C3 2s2 + s + 1 = ; 2 s 1 s +s+1 (s 1)(s2 + s + 1) 4 C2 + C3 ) + ( C3 ) = 2s2 + s + 1; 3 4 2 + C2 = 2 ) C2 = ; 3 3 4 1 C3 = 1 ) C3 = : 3 3

4 4 s2 ( + C2 ) + s( 3 3

F (s)

=

=

4 3 s 4 3

+

2 1 s+ 3 ; + 3 1 s2 + s + 1 2 + 1 3

s+

1 2

p !2 ; 3 + 2 p 4 t 2 t 3 1 f (t) = L fF (s)g = e + e 2 cos t; 3 3 2 ( p ) t 2 3 2et + e 2 cos t 1(t): = 3 2 s

1 s+ 2

2

We can verify the answer using Matlab: >> ilaplace((2*s^2+s+1)/(s^3-1)) ans = 4/3*exp(t)+2/3*exp(-1/2*t)*cos(1/2*3^(1/2)*t)

3018

CHAPTER 3. DYNAMIC RESPONSE (c) Carry out partial fraction expansion, F (s)

= =

C1

=

C3

=

C2

= = = =

2(s2 + s + 1)

2 ; s (s + 1) C1 C2 C3 : + + s (s + 1) (s + 1)2 2(s2 + s + 1) sF (s)js=0 = js=0 = 2; 2 (s + 1) 2(s2 + s + 1) (s + 1)2 F (s)js= 1 = js= s d [(s + 1)2 F (s)]s= 1 ; ds d 2(s2 + s + 1) [ ]s= 1 ; ds s 2(2s + 1)s 2(s2 + s + 1) js= 1 ; s2 0:

1

=

2;

2 0 2 + + : s (s + 1) (s + 1)2

F (s)

=

f (t)

= L

1

fF (s)g = 2f1

te t g1(t):

We can verify the answer using Matlab: >> ilaplace((2*s^2+2*s+2)/(s*(s+1)^2)) ans = 2-2*t*exp(-t) (d) Carry out partial fraction expansion, F (s)

= = =

1 1 3 1 s3 + 2s + 4 As + B Cs + D 4s + 2 4s 2 = + = + ; s4 16 s2 4 s2 + 4 s2 4 s2 + 4 3 d 1 1 1 d 1 1 sinh(2t) + sinh(2t) sin(2t) sin(2t) ; 4 4 dt 2 4 4 dt 2 1 3 1 1 sinh(2t) + cosh(2t) sin(2t) + cos(2t): 4 4 4 4

We can verify the answer using Matlab: >> ilaplace((s^3+2*s+4)/(s^4-16)) ans = -1/4*sin(2*t)+1/2*exp(2*t)+1/4*exp(-2*t)+1/4*cos(2*t) (e) Expand in partial fraction expansion and compute the residues using

3019 the results from Appendix A, F (s)

= =

C1

=

C4

=

C5

=

C2

=

C3

= =

2(s + 2)(s + 5)2 ; (s + 1)(s2 + 4)2 C1 C5 C2 C3 C4 + : + + + s + 1 s 2j s + 2j (s 2j)2 (s + 2j)2 32 (s + 1)F (s)js= 1 = = 1:280; 25 83 39j = 4:150 j1:950; (s 2j)2 F (s)js=2j = 20 C4 = 4:150 + j1:950; d 128 579j (s 2j)2 F (s) s=2j = ; ds 200 0:64 j2:895; C2 = 0:64 + j2:895:

These results can also be veri…ed with the Matlab residue command, a =[1 1 8 8 16 16]; b =[2 24 90 100]; [r,p,k]=residue(b,a) r= -0.64000000000000 - 2.89500000000002i -4.15000000000002 - 1.95000000000000i -0.64000000000000 + 2.89500000000002i -4.15000000000002 + 1.95000000000000i 1.28000000000001 p= 0.00000000000000 + 2.00000000000000i 0.00000000000000 + 2.00000000000000i 0.00000000000000 - 2.00000000000000i 0.00000000000000 - 2.00000000000000i -1.00000000000000 k= [] We then have, f (t)

= 1:28e = 1:28e

t t

+ 2jC2 j cos(2t + arg C2 ) + 2jC4 jt cos(2t + arg C4 ); + 5:92979 cos(2t 1:788) + 9:1706t cos(2t 2:702):

where jC2 j = 2:96489;

jC4 j = 4:5853;

arg C2 = tan

1

2:895 0:64

=

1:788;

3020

CHAPTER 3. DYNAMIC RESPONSE using the atan2 command in Matlab, and arg C4 = tan

1:950 4:150

1

=

2:702;

also using the atan2 command in Matlab. (f) (s2 1) : (s2 + 1)2

F (s) =

Using the multiplication by time Laplace transform property (Table A.1 entry #11): d G(s) = Lftg(t)g: ds We can see that s2 1 d s : = 2 2 ds (s + 1) (s + 1)2 So the inverse Laplace transform of F (s) is: L

1

fF (s)g = t cos t:

We can verify the answer using Matlab: >> ilaplace((s^2-1)/(s^2+1)^2) ans = t*cos(t) (g) Follows from Problem 5 (c), or expand in series, 1

tan

1 1 ( )= s s

1 1 + 5 3 3s 5s

:::

Then, L

1

tan

1

1 ( ) s

t2 t4 + 3! 5!

=1

::::: =

sin(t) : t

Alternatively, let us assume L

1

tan

1

1 ( ) s

= f (t):

We use the identity d tan ds

1

s =

1 ; 1 + s2

which means that L

1

1 s2 + 1

=

tf (t) =

sin(t):

3021 Therefore, sin(t) : t We can verify the answer using Matlab: >> ilaplace(atan(1/s)) ans = 1/t*sin(t) f (t) =

9. Solve the following ordinary di¤erential equations using Laplace transforms: (a) y•(t) + y(t) _ + 3y(t) = 0; y(0) = 1; y(0) _ =2 (b) y•(t)

2y(t) _ + 4y(t) = 0; y(0) = 1; y(0) _ =2

(c) y•(t) + y(t) _ = sin t; y(0) = 1; y(0) _ =2 (d) y•(t) + 3y(t) = sin t; y(0) = 1; y(0) _ =2 (e) y•(t) + 2y(t) _ = et ; y(0) = 1; y(0) _ =2 (f) y•(t) + y(t) = t; y(0) = 1; y(0) _ =

1

Solution: (a) y•(t) + y(t) _ + 3y(t) = 0;

y (0) = 1;

y_ (0) = 2

Using Table A.1 entry #5, the di¤erentiation Laplace transform property, s2 Y (s)

Y (s)

sy (0)

= =

s2

y_ (0) + sY (s)

s+3 ; +s+3 s + 21 + 52

s+ = s

1 2 2

+

11 4

s + 12 2 + 12 + 11 4

y (0) + 3Y (s) = 0

; p 5 11 + 11 s +

Using Table A.2 entries #19 and #20, p p 1 11 5 11 t 2 cos t+ e y(t) = e 2 11

1 2t

q

11 4 1 2 + 11 2 4

sin

:

p

11 t: 2

We can verify the answer using Matlab: >> dsolve(’D2y+Dy+3*y=0’,’y(0)=1’,’Dy(0)=2’,’t’) ans = 5/11*11^(1/2)*exp(-1/2*t)*sin(1/2*11^(1/2)*t)+exp(-1/2*t)*cos(1/2*11^(1/2)*t)

3022

CHAPTER 3. DYNAMIC RESPONSE (b) y•(t) 2

s Y (s)

2y(t) _ + 4y (t) = 0; y (0) = 1; y_ (0) = 2: sy (0)

y_ (0)

Y (s)

2sY (s) + 2y (0) + 4Y (s) = 0:

= =

s2

s ; 2s + 4 s

(s

1) + 3

2

;

Using Table A.1 entry #5 and Table A.2 entry #20, p i d h t y(t) = e sin 3t dt p p 1 y(t) = p et sin 3t + et cos 3t 3 We can verify the answer using Matlab: >> dsolve(’D2y-2*Dy+4*y=0’,’y(0)=1’,’Dy(0)=2’,’t’) ans = 1/3*3^(1/2)*exp(t)*sin(3^(1/2)*t)+exp(t)*cos(3^(1/2)*t) (c) y•(t) + y(t) _ = sin t; s2 Y (s)

sy (0)

Y (s)

= =

s3

3 + C3 2

C1

=

C2

=

y (0) = 1;

y_ (0) + sY (s)

y_ (0) = 2 1 y (0) = 2 s +1

s3 + 3s2 + s + 4 ; s (s + 1) (s2 + 1) C1 C2 C3 s + C4 + + 2 : s s+1 s +1

s3 + 3s2 + s + 4 js=0 = 4; (s + 1) (s2 + 1) 5 s3 + 3s2 + s + 4 js= 1 = : s (s2 + 1) 2

5 4 C3 s + C4 s3 + 3s2 + s + 4 2 + + 2 = s s+1 s +1 s (s + 1) (s2 + 1) 3 + s2 (4 + C3 + C4 ) + s + C4 + 4 = s3 + 3s2 + s + 4: 2

Match coe¢ cients of like powers of s 3 2 3 C3 + 2 C4 +

=

1

=) C4 =

=

1

=) C3 =

1 ; 2 1 : 2

3023 5 1 5 s 12 4 4 1 s 2 2 + + 22 = + s s+1 s +1 s s + 1 2 s2 + 1 Using Table A.2 entries #2, #7, #17, and #18

5 e 2

y (t) = 4

t

1 cos t 2

1 1 : 2 s2 + 1

1 sin t: 2

We can verify the answer using Matlab: >> dsolve(’D2y+Dy-sin(t)=0’,’y(0)=1’,’Dy(0)=2’,’t’) ans = -1/2*sin(t)-1/2*cos(t)-5/2*exp(-t)+4 (d) y• (t) + 3y (t) = sin t; s2 Y (s)

sy (0)

Y (s)

= =

y (0) = 1;

y_ (0) = 2;

y_ (0) + 3Y (s) =

1 ; s2 + 1

s3 + 2s2 + s + 3 ; (s2 + 3) (s2 + 1) C3 s + C4 C1 s + C2 + 2 : 2 s +3 s +1

(C1 s + C2 ) s2 + 1 + (C3 s + C4 ) s2 + 3 s3 + 2s2 + s + 3 = 2 : 2 2 (s + 3) (s + 1) (s + 3) (s2 + 1) Match coe¢ cients of like powers of s: s3 (C1 + C3 )+s2 (C2 + C4 )+s (C1 + 3C3 )+(C2 + 3C4 ) = s3 +2s2 +s+3; C1 + C3 C2 + C4 C1 + 3C3

= = = =)

C2 + 3C4

=

1 =) C1 = C3 + 1; 2 =) C2 = 2 C4 ; 1 =) C3 + 1 + 3C3 = 1 C1 = 1; 3

=) (2

=) C2 =

C4 ) + 3C4 = 3

=) C3 = 0; =) C4 =

3 ; 2 1 3 2s + 2 s2 + 3

=

+

1 2

; s2 + 1 p p 1 3 3 1 1 2 = + + : s2 + 3 2 s2 + 3 2 s2 + 1 p p p 1 3 1 y (t) = cos 3t + sin 3t + sin t: 2 2 2

Y (s)

1 ; 2

3024

CHAPTER 3. DYNAMIC RESPONSE We can verify the answer using Matlab: >> dsolve(’D2y+3*y-sin(t)=0’,’y(0)=1’,’Dy(0)=2’,’t’) ans = 1/2*sin(3^(1/2)*t)*3^(1/2)+cos(3^(1/2)*t)+1/2*sin(t) (e) y•(t) + 2y_ (t) = et ; s2 Y (s)

sy (0)

Y (s)

y_ (0) + 2sY (s)

=

C2

=

C3

=

y_ (0) = 2 2y (0) =

s2 + 3s 3 js=0 = (s 1) (s + 2) s2 + 3s 3 1 js=1 = ; s (s + 2) 3 2 s + 3s 3 js= 2 = s (s 1)

Y (s) =

1 s

s2 + 3s 3 ; s (s 1) (s + 2) C1 C2 C3 + + : s s 1 s+2

= =

C1

y (0) = 1;

3 2

+

s

1 1 3s 1

3 1 t + e 2 3

y (t) =

3 ; 2

5 ; 6

5 1 : 6s+2 5 e 6

2t

:

We can verify the answer using Matlab: >> dsolve(’D2y+2*Dy-exp(t)=0’,’y(0)=1’,’Dy(0)=2’,’t’) ans = 1/3*exp(t)-5/6*exp(-2*t)+3/2 (f) Using the results from Appendix A, y• (t) + y (t) = t; s2 Y (s)

Y (s)

sy (0)

= =

y (0) = 1;

y_ (0) =

y_ (0) + Y (s) =

1 ; s2

s3 s2 + 1 ; s2 (s2 + 1) C1 C2 C3 s + C4 + 2 + 2 : s s s +1

1;

1

3025 C1

=

C2

=

d s3 s2 + 1 js=0 = 0; ds (s2 + 1) s3 s2 + 1 js=0 = 1: (s2 + 1)

C3 s + C4 1 + 2 2 s s +1 2 s + 1 + (C3 s + C4 ) s2 s2 (s2 + 1)

=

s3 s2 + 1 ; s2 (s2 + 1)

=

s3 s2 + 1 : s2 (s2 + 1)

Match coe¢ cients of like powers of s: C3 = 1 C4 + 1 = 1 Y (s) =

=) C4 =

s 1 + 2 2 s s +1

y (t) = t + cos t

2

s2

2

1 +1

2 sin t:

We can verify the answer using Matlab: >> dsolve(’D2y+y-t=0’,’y(0)=1’,’Dy(0)=-1’,’t’) ans = -2*sin(t)+cos(t)+t 10. Using the convolution integral, …nd the step response of the system whose impulse response is given below and shown in Figure 3.47: te t 0

h(t) =

t 0 t < 0:

Solution: There are only two cases to consider. Case (a): For the case t 0, the situation is illustrated in the following Figure part (c). There is no overlap between the two functions (u(t ) and h( )) and the output is zero y1 (t) = 0: Case (b): For the case t 0, the situation is displayed in the following Figure part (d). The output of the system is given by

y2 (t) =

Zt 0

h( )u(t

)d =

Zt 0

( e

)(1)d = 1

(t + 1)e t :

3026

CHAPTER 3. DYNAMIC RESPONSE

Figure 3.47: Impulse response for Problem 3.10.

Illustration of convolution. The output of the system is the composite of the two segments computed

3027

Figure 3.48: Impulse response for Problem 3.11.

above as shown in the following Figure.

System output response. 11. Using the convolution integral, …nd the step response of the system whose impulse response is given below and shown in Figure 3.48: h(t) =

1 0 t 2 0 t < 0 and t > 2

Solution: There are three cases to consider as shown in the following …gure. Case (a): For the case t 0, the situation is illustrated in the following Figure part (c). There is no overlap between the two functions (u(t ) and h( )) and the output is zero y1 (t) = 0 Case (b): For the case 0 > t 2, the situation is displayed in the following Figure part (d) and shows partial overlap. The output of the system is

3028

CHAPTER 3. DYNAMIC RESPONSE given by

y2 (t) =

Zt

h( )u(t

)d =

0

Zt

(1)(1)d = t:

0

Case (c): For the case t 2, the situation is displayed in the following Figure part (e) and shows total overlap. The output of the system is given by

y3 (t) =

Zt 0

h( )u(t

)d =

Z2 0

(1)(1)d = 2:

3029

Illustration of convolution.

The output of the system is the composite of the three segments computed above as shown in the following …gure.

3030

CHAPTER 3. DYNAMIC RESPONSE

System output response. 12. Consider the standard second-order system G(s) =

! 2n : s2 + 2 ! n s + ! 2n

a) Write the Laplace transform of the signal in Fig. 3.49. b). What is the transform of the output if this signal is applied to G(s). c) Find the output of the system for the input shown in Fig. 3.49. u ( t) 1

1

2

3

T im e (s e c )

Figure 3.49: Plot of input for Problem 3.12 Solution: (a) The input signal in Figure 3.40 may be written as: u(t) = t

t[1(t

1)]

t[1((t

2)] + t[1(t

3)];

where 1(t ) denotes a delayed unit step. The Laplace transform of the input signal is: U (s) =

1 1 s2

e

s

e

2s

e

3s

:

(b) The Laplace transform of the output if this signal is applied is:

Y (s) = G(s)U (s) =

! 2n s2 + 2 ! n s + ! 2n

1 s2

1

e

s

e

2s

e

3s

:

3031 (c) However to make the mathematical manipulation easier, consider only the response of the system to a ramp input: Y1 (s) =

! 2n + 2 ! n s + ! 2n

s2

1 s2

:

Partial fractions yields the following: 2 !n

1 Y1 (s) = 2 s

s

2 !n

+

!n 2

s + 2 !n

(s + ! n )2 + ! n

p 1

2

2:

Use the following Laplace transform pairs for the case 0

L

1

(

s + z1 2

(s + a) + ! 2

)

=

s

2

a) + ! 2 e !2

(z1

where

L

1

z1

1 s2

1 s

1

L

!

1

tan

sin(!t + );

:

a

=t

at

ramp

= 1(t)

unit step

and the following Laplace transform pairs for the case L L

1

(

L L

1

1

(

1 2

(s + a) s 2

(s + a) 1

1 s

1 s2

)

)

= te

= (1

=t

= 1(t)

> < t !2n + e p 2 sin ! n 1 !n 1 y1 (t) = > 2 > t !n + !2n e !n t !2n t + 1 :

2

t + tan

1 2

p 2

2

1

2

1

0

0. Solution: K K2 )2 ; F2 = 1 + KH1 1 + K 2 H2 2 2 F2 = ; SK = 1 + KH1 1 + K 2 H2

F1 = ( F1 SK

F1 = F2 =) H2 = H12 + F2 SK =

2H1 K

F1 SK 2 = (1 + KH1 )2 1 + KH1

System 2 is less sensitive. The conclusion is to put as much gain in the feedback loop as you can. 4. A unity feedback control system has the open-loop transfer function G(s) =

A : s(s + a)

(a) Compute the sensitivity of the closed-loop transfer function to changes in the parameter A.

4003 (b) Compute the sensitivity of the closed-loop transfer function to changes in the parameter a. (c) If the unity gain in the feedback changes to a value of 6= 1, compute the sensitivity of the closed-loop transfer function with respect to . Solution: (a) G(s) = T (s) = 1 + G(s)

A A s(s + a) = 2 A s + as + A 1+ s(s + a)

dT (s2 + as + A) A = 2 dA (s2 + as + A) T SA =

A(s2 + as + A) s2 + as s(s + a) A dT = = 2 2 T dA A (s + as + A) s(s + a) + A

(b) dT sA = 2 da (s + as + A)2 a(s2 + as + A) sA a dT = T da A (s2 + as + A)2 SaT =

as s(s + a) + A

(c) In this case, T (s) =

G(s) 1 + G(s)

dT G(s)2 = d (1 + G(s))2 dT = T d

(1 + G) G2 G = G (1 + G)2 1+ G

A A s(s + A) S = = A s(s + a) + A 1+ s(s + a) T

If a = A = 1;the transfer function is most sensitive to variations in a and A near ! = 1 rad/sec . The steady-state response is not a¤ected by variations in A and T a (SA (0) and SaT (0) are both zero). The steady-state response is heavily dependent on since jS T (0)j = 1.0

4004 See attached plots of sensitivities versus radian frequency for a = A = 1:0: 1

10

sensitivity to A

0

10

sensitivity to a

sensitivity to beta -1

10

-2

10

-3

10

0

1

2

3

4

5

6

7

8

9

10

5. Compute the equation for the system error for the …ltered feedback system shown in Fig. 4.4. Solution: For this …gure, the equation for the output is: Y =

F DG G R+ W 1 + DGH 1 + DGH

DGH V 1 + DGH

And the resulting equation for the error is: E

= R Y 1 + DG(H F ) R = 1 + DGH

G DGH W+ V 1 + DGH 1 + DGH

Therefore, as we have seen, increasing the loop gain does not necessarily reduce the error as the result depends on the structure of the system.. 6. If S is the sensitivity of the …ltered feedback system to changes in the plant transfer function and T is the transfer function from reference to output, compute the sum of S + T . Show that S + T = 1 if F = H: (a) Compute the sensitivity of the …ltered feedback system shown in Fig 4.4 with respect to changes in the plant transfer function, G: (b) Compute the sensitivity of the …ltered feedback system shown in Fig 4.4 with respect to changes in the controller transfer function, Dcl: (c) Compute the sensitivity of the …ltered feedback system shown in Fig 4.4 with respect to changes in the …lter transfer function, F:

4005 (d) Compute the sensitivity of the …ltered feedback system shown in Fig 4.4 with respect to changes in the sensor transfer function, H: Solution: To answer the …rst question, we need the answer to part a) so let’s start there. a. Applying the formula for sensitivity of T to changes in G: T S

F DG 1 + DGH 1 + DGH (1 + DGH)F D (F DG)(DH = G F DG (1 + DGH)2 1 = 1 + DGH =

Now we can do S+T

= = =

1 F DG + 1 + DGH 1 + DGH 1 + F DG 1 + DGH 1 if F = H

(4.1)

b. Applying the formula for sensitivity of T to changes in D: STD

1 + DGH (1 + DGH)F G F DG(GH) F DG (1 + DGH)2 1 1 + DGH

= D =

This is not surprising as D and G are in series. c. Applying the formula for sensitivity of T to changes in F: STF

1 + DGH (1 + DGH)(DG) F DG (1 + DGH)2 1 + DGH 1 + DGH 1

= F

=

In this case, the F term is in the open loop so it has sensitivity of unity. d. Applying the formula for sensitivity of T to changes in H: STH

= H =

1 + DGH (1 + DGH)0 F DG(DG) F DG (1 + DGH)2 DGH (1 + DGH)

4006

Figure 4.25: Control system for Problem 7

Figure 4.26: Control system for Problem 8

Problems and Solutions for Section 4.2: Control of Steady-State Error 7. Consider the DC-motor control system with rate (tachometer) feedback shown in Fig. 4.25(a). (a) Find values for K 0 and kt0 so that the system of Fig. 4.25(b) has the same transfer function as the system of Fig. 4.25(a). (b) Determine the system type with respect to tracking the system Kv in terms of parameters K 0 and kt0 .

r

and compute

(c) Does the addition of tachometer feedback with positive kt increase or decrease Kv ? Solution: (a) Using block diagram reduction techniques: 1 to its output. k - Eliminate the second summer by absorbing Kp : This will result in Figure 4.25(b) where - Move the picko¤ point from the input of the

K0 =

Kp KKm k

kt0 =

kkt : Kp

4007 (b) The inner-loop in Fig. 4.25(a) may be reduced to kKm + kKm kt )

s(1 +

ms

which means that the unity feedback system has a pure integrator in the forward loop and hence it is Type 1 with respect to reference kKm input ( r ) and Kv = (1 + kKm kt ) (c) We conclude that the introduction of kt reduces the velocity constant and therefore makes the error to a ramp larger 8. Consider the system shown in Fig. 4.26, where D(s) = K

(s + )2 : s2 + ! 2o

(a) Prove that if the system is stable, it is capable of tracking a sinusoidal reference input r = sin ! o t with zero steady-state error. (Look at the transfer function from R to E and consider the gain at ! o :) (b) Use Routh’s criterion to …nd the range of K such that the closed-loop system remains stable if ! o = 1 and = 0:25: Solution: (a) D(s)G(s)

=

E(s) R(s)

=

K(s + )2 + ! 2o )s(s + 1) 1 1 + DG (s2

2

=

s(s + 1)(s + ! 2o ) 2

(s2 + ! 2o )s(s + 1) + K(s + )

The gain of this transfer function is zero at s = j! o and we expect the error to be zero if R is a sinusoid at that frequency. More formally, !n let R(s) = 2 then s + ! 2n 2

E(s) =

s(s + 1)(s + ! 2o ) (s2

+

! 2o )s(s

2

+ 1) + K(s + )

!n s2 + ! 2n

Assuming the (closed-loop) system is stable, then if ! n : ! o Then E(s) has a pole on the imaginary axis and the FVT does not apply. The …nal error will NOT be zero in this case. However, if ! n = ! o we can use the FVT and ess = lim sE(s) = 0 s!0

4008

Figure 4.27: Control system for problem 9

(b) To test for stability, the characteristic equation is, s4 + (K + ! 2o )s2 + s3 + (! 2o + 2 K)s + K

2

=0

Using the Routh array s4 : s3 : s2 :

! 2o + K K (! 2o + 2 K) K 2

1 1 K(1

2 )

2

2 1

s :

! 2o

+2 K +

s0 : If

K

(1

2 )

2

= 0:25; we must have K > 0, and K > 0:25

2! 2o .

9. Consider the system shown in Fig. 4.27 which represents control of the angle of a pendulum that has no damping. (a) What condition must D(s) satisfy so that the system can track a ramp reference input with constant steady-state error? (b) For a transfer function D(s) that stabilizes the system and satis…es the condition in part (a), …nd the class of disturbances w(t) that the system can reject with zero steady-state error. Solution: (a) For a unity feedback system to be Type 1 the open loop transfer function must have a pole at s = 0: Thus in this case, since G has no such pole, it is necessary for D to have a pole at s = 0: (b) Y (s) = lim s(

s!0

1 W (s) s2 + D(s) + K

1 1 ) =0 s2 + D(s) + K s`

4009 i¤ lim s`

s!0

1

D(s) = 1

i¤ ` = 1 since D(s) has a pole at the origin. Therefore system will reject step disturbanceswith zero error. 10. A unity feedback system has the overall transfer function ! 2n Y (s) : = T (s) = 2 R(s) s + 2 ! n s + ! 2n Give the system type and corresponding error constant for tracking polynomial reference inputs in terms of and ! n . Solution:

E(s) s2 + 2 ! n s = 2 R(s) s + 2 ! n s + ! 2n

Therefore the system is Type 1 and the velocity constant is Kv =

!n 2

11. Consider the second-order system G(s) =

1 : s2 + 2 s + 1

We would like to add a transfer function of the form D(s) = K(s+a)=(s+b) in series with G(s) in a unity-feedback structure. (a) Ignoring stability for the moment, what are the constraints on K; a, and b so that the system is Type 1? (b) What are the constraints placed on K, a, and b so that the system is both stable and Type 1? (c) What are the constraints on a and b so that the system is both Type 1 and remains stable for every positive value for K? Solution: (a) In a unity feedback structure, the error is 1=(1+GD) and, as we saw, to be Type 1, there needs to be a pole at s = 0 in the product GD: Since there is no such pole in G; it must be supplied by D; thus, the answer is b=0 (b) To assure stability, all poles of the closed loop must be in the left half plane, for which the criterion is by Routh. Thus the characteristic equation is s(s2 + 2 s + 1) + K(s + a) = 0

4010 and the Routh array is 1 2 2 (1 + K) 2 aK

1+K aK aK

Thus the requirements are 2 (1 + K)

aK aK

> >

0 0

(c) If we assume that > 0 and, for this part, that a > 0 also, the requirements can be reduced to

2 + K(2

K > 0 a) > 0

If a < 2 ; inspection of these conditions shows that the system will be stable for all positive values of K: On the other hand, if a > 2 ; then the requirement is 0 0 and 11(1 + kp ) (b) From above, with = 0:9 E(s)

= =

10(1 + kp ) 10kI

kI > 0 for stability.

s(s + 1)(s + 10) + 9(kp s + kI ) 10(kp s + kI ) ]R(s) s(s + 1)(s + 10) + 9(kp s + kI ) s(s + 1)(s + 10) k p k I R(s) s(s + 1)(s + 10) + 9(kp s + kI )

[

4023

Figure 4.36: Control system for Problem 21

If kp and kI are chosen so that the system is stable, applying the 1 FVT for R(s) = 2 results in ess ! 1: The system is no longer s Type 1. 1 (c) Try R(s) = s lim sE(s)

s!0

= =

lim

s(s + 1)(s + 10)

kp s kI

s(s + 1)(s + 10) + 9(kp s + kI ) kI 1 = 9kI 9

s!0

1 . 1 + Kp Thus, Kp = 8: Without the magnitude an equivalent result is that Kp = 10: and the system is Type 0. Kp is de…ned such that jess j =

21. Consider the system shown in Fig. 4.36. (a) Find the transfer function from the reference input to the tracking error. (b) For this system to respond to inputs of the form r(t) = tn 1(t) (where n < q) with zero steady-state error, what constraint is placed on the open-loop poles p1 ; p2 ; ; pq ? Solution: (a)

(b)

Qq (s + pi ) 1 E(s) = = Qq i=1 R(s) 1 + G(s) i=1 (s + pi ) + 1 r(t) = tn =) R(s) = ess

n! sn+1

Qq (s + pi ) = lim s n+1 Qq i=1 s!0 s i=1 (s + pi ) + 1 n!

If ess is to be zero the system must have at least n + 1 poles at

4024 the origin: ess

Qq sn+1 i=1 (s + pi ) =0 = lim s n+1 n+1 Qq s!0 s s i=1 (s + pi ) + 1 n!

22. A linear ODE model of the DC motor with negligible armature inductance (La = 0) and with a disturbance torque w was given earlier in the chapter; it is restated here, in slightly di¤erent form, as JRa • _ m + Ke m = Kt

a

+

Ra w; Kt

where m is measured in radians. Dividing through by the coe¢ cient of •m , we obtain •m + a1 _ m = b0 a + c0 w; where

Kt 1 Kt Ke ; b0 = ; c0 = : JRa JRa J With rotating potentiometers, it is possible to measure the positioning error between and the reference angle r or e = ref m . With a tachometer we can measure the motor speed _ m . Consider using feedback of the error e and the motor speed _ m in the form a1 =

a

= K(e

TD _ m );

where K and TD are controller gains to be determined. (a) Draw a block diagram of the resulting feedback system showing both _ m and m as variables in the diagram representing the motor. (b) Suppose the numbers work out so that a1 = 65, b0 = 200, and c0 = 10. If there is no load torque (w = 0), what speed (in rpm) results from va = 100 V? (c) Using the parameter values given in part (b), let the control be D = kp + kD s and …nd kp and kD so that, using the results of Chapter 3, a step change in ref with zero load torque results in a transient that has an approximately 17% overshoot and that settles to within 5% of steady-state in less than 0:05 sec. (d) Derive an expression for the steady-state error to a reference angle input, and compute its value for your design in part (c) assuming ref = 1 rad. (e) Derive an expression for the steady-state error to a constant disturbance torque when ref = 0, and compute its value for your design in part (c) assuming w = 1:0. Solution: (a) Block diagram:

4025

Block diagram for Problem 4.22 (b) If Va =constant the system is in steady state: _ = b0 Va = 200 100 60 rad:sec a1 65 2 rpm

1

= 2938 rpm

(c) = r

s2

Kb0 ! 2n = 2 + s(a1 + TD Kb0 ) + Kb0 s + 2 ! n s + ! 2n

Mp = 17%; =):

= 0:5 ts = 0:05 sec. to 5% : ! n ts

=) e

= 0:05 =) ! n = 60 =) ! n = 120

Comparing coe¢ cients: K = 72 ;

TD = 3:8

10

3

(d) Steady-state error: E(s) = For

r

=

1 : s

=

r

s(s + a1 + TD Kb0 ) s2 + s(a1 + TD Kb0 ) + Kb0

ess = lim sE(s) = 0 s!0

r

(Type 1)

(e) Response to torque:

QL ss

=

c0 s2 + s(a1 + TD Kb0 ) + Kb0

= lim s: (s) = lim s s!0

s!0

c0 1 c0 1 = = rad s2 + : : : s Kb0 1440

4026 23. We wish to design an automatic speed control for an automobile. Assume that (1) the car has a mass m of 1000 kg, (2) the accelerator is the control U and supplies a force on the automobile of 10 N per degree of accelerator motion, and (3) air drag provides a friction force proportional to velocity of 10 N sec=m. (a) Obtain the transfer function from control input U to the velocity of the automobile. (b) Assume the velocity changes are given by V (s) =

0:05 1 U (s) + W (s); s + 0:02 s + 0:02

where V is given in meters per second, U is in degrees, and W is the percent grade of the road. Design a proportional control law U = kp V that will maintain a velocity error of less than 1 m/sec in the presence of a constant 2% grade. (c) Discuss what advantage (if any) integral control would have for this problem. (d) Assuming that pure integral control (that is, no proportional term) is advantageous, select the feedback gain so that the roots have critical damping ( = 1). Solution: a. X m• x = F = Ka u Dx_ Lfmv_ V U

b.

= Ka u Dvg Ka 0:01 = = ms + D s + 0:01

Error: 1 kp 0:05 s + 0:02 s + 0:02 G(s) Vd + E(s) = Vd V = Vd kp kp 1+ 1+ s + 0:02 s + 0:02 (s + 0:02)Vd 0:05G = s + 0:02 + kp

If we want error < 1 m/sec in presence of grade, we in fact need jess (step)j < 1. Assume no input : (Vd = 0) ess (step) = lim s( s!0

j

0:05 2 0:1 ) = s + 0:02 + kp s 0:02 + kp

0:1 j 0:08 c. The obvious advantage of integral control would be zero s.s. error for step input (Type 1 system would result). kI d. Pure integral control: kp ! s E(s) =

s(s + 0:02)Vd 0:05sG(s) s2 + 0:02s + kI

= 1 =) ! n = 0:01 =) kI = 0:0001 24. Consider the automobile speed control system depicted in Fig. 4.37. (a) Find the transfer functions from W (s) and from R(s) to Y (s). (b) Assume that the desired speed is a constant reference r, so that R(s) = ro =s. Assume that the road is level, so w(t) = 0. Compute values of the gains K, Hr , and Hf to guarantee that lim y(t) = ro :

t!1

Include both the open-loop (assuming Hy = 0) and feedback cases (Hy 6= 0) in your discussion.

(c) Repeat part (b) assuming that a constant grade disturbance W (s) = wo =s is present in addition to the reference input. In particular, …nd the variation in speed due to the grade change for both the feed forward and feedback cases. Use your results to explain (1) why feedback control is necessary and (2) how the gain kp should be chosen to reduce steady-state error.

4028 (d) Assume that w(t) = 0 and that the gain A undergoes the perturbation A + A. Determine the error in speed due to the gain change for both the feed forward and feedback cases. How should the gains be chosen in this case to reduce the e¤ects of A? Solution: (a) Y (s) =

kp AHr B W (s) + R(s) s + a + kp AHy s + a + kp AHy

(b) Feedforward:(Hy = 0) lim sY (s) = lim

s!0

s!0

therefore, kp =

kp AHr r=r s+a+0

a : AHr

Feedback: lim y(t) = r

t!1

results in

kp AHr r=r a + kp AHy Choose kp for performance and Hy for sensor characteristics, and set Hr =

a + Akp Hy kp A

(c) Feedforward: Bw ar + a a Bw =r+ a

lim y(t) =

t!1

Therefore, Bw ; a all quantities are …xed- no way to reduce e¤ect of disturbance. Feedback: kp AHr B w+ r lim y(t) = t!1 a + kp AHy a + kp AHy yf f (1) =

=

B w+r a + kp AHy

(if Hr is chosen as in part (b)). Therefore, yf b (1) =

B w a + kp AHy

E¤ect of disturbance can be made small by choosing kp large.

4029 (d) Feedforward: using kp =

a as derived in part (b), AHr

yf f (1) = (1 +

A )r A

therefore, yf f (1) =

A r A

or

A yf f (1) = r A which means that 5% error in A results in 5% error in tracking. Feedback: (A + A)kp Hr yf b (1) = r a + (A + A)kp Hy

using value for Hr chosen in part (b) gives yf b (1) = [

(A + A) a + kp AHy ]r a + (A + A)kp Hy A

=r+

a A r aA + (A + A)kp AHy

=r+

A a a + kp AHy A

a yf b (1) A = r a + kp AHy A Tracking error due to parameter variation can be reduced by choosing kp large. 25. Consider the multivariable system shown in Fig. 4.38. Assume that the system is stable. Find the transfer functions from each disturbance input to each output and determine the steady-state values of y1 and y2 for constant disturbances. We de…ne a multivariable system to be Type k with respect to polynomial inputs at wi if the steady-state value of every output is zero for any combination of inputs of degree less than k and at least one input is a non-zero constant for an input of degree k: What is the system type with respect to disturbance rejection at w1 ? At w2 ? Solution: (a) 1 s s(s + 1) R1 + 2 W1 + 2 W2 s2 + s + 1 s +s+1 s +s+1 W10 W20 ; W2 (s) = For constant disturbances, R1 = 0; W1 (s) = s s Y1 =

Y1 =

W10 + (s + 1)W20 s2 + s + 1

4030

Figure 4.38: Multivariable system

Let u2 be the signal coupling systems 1 and 2: U2

=

Y2

= =

(s + 1)(R1

W2 ) + s(s + 1)W1 s2 + s + 1 R2 (s + 1)U2 + 2 2 s + 3s + 2 s + 3s + 2 (s + 1)2 ( W2 ) + s(s + 1)2 W2 (s2 + 3s + 2)(s2 + s + 1)

The system type w.r.t. disturbances: y1 w:r:t: W1

Type 1

y1 w:r:t: W2

Type 1

y2 w:r:t: W1

Type 1

y2 w:r:t: W2

Type 0

can be determined by applying FVT to Y1 and Y2 or by inspection. Problems and Solutions for Section 4.3: The Three Term controller: PID control 26. The transfer functions of speed control for a magnetic tape-drive system are shown in Fig. 4.39. The speed sensor is fast enough that its dynamics can be neglected and the diagram shows the equivalent unity feedback system. (a) Assuming the reference is zero, what is the steady-state error due to a step disturbance torque of 1 N m? What must the ampli…er gain K be in order to make the steady-state error ess 0:001 rad/sec.? (b) Plot the roots of the closed-loop system in the complex plane, and accurately sketch the time response of the output for a step reference input using the gain K computed in part (a).

4031

Figure 4.39: Speed-control system for a magnetic tape drive

(c) Plot the region in the complex plane of acceptable closed-loop poles corresponding to the speci…cations of a 1% settling time of ts 0:1 sec. and an overshoot Mp 5%. (d) Give values for kp and kD for a PD controller which will meet the speci…cations. (e) How would the disturbance-induced steady-state error change with the new control scheme in part (d)? How could the steady-state error to a disturbance torque be eliminated entirely? Solution: (a) TF for disturbance:

Y = W

1 Js + b 1 10kp 1+ : Js + b 0:5s + 1

ess (step in W) = lim s s!0

ess

0:01 ; kp

b = 1 ; J = 0:1

1Y 1 = sW 1 + 10kp

9:9 pick kp = 10:

(b) Y (s) = r (s)

1 10kp : 2000 0:5s + 1 Js + b = 2 1 10kp s + 12s + 2020 1+ : Js + b 0:5s + 1

!n =

p

2020 ' 45 ;

12 = p = 0:13 2 2020

The roots are undesirable (damping too low, high overshoot).

4032 50

Linesofdampingandwn

stepresponse

*

1.8

40 1.6

30 1.4

20

1.2 Amplitude

10 0

1

-10

0.8

-20

0.6

-30

0.4 0.2

-40 * -50 -50

(c) For ts

-40

-30

0:1 =)

-20

-10

0

0

46 For Mp

0

0.5 Time(sec)

0:05 =)

1

0:7

Lines of damping and wn 60

40

20

0

-20

-40

-60 -160

-140

-120

-100

-80

-60

-40

-20

0

s-plane for part(c) (a) d. We know that larger ! n and are needed. This can be achieved by increasing kp and adding derivative feedback as in Fig. 4.12 ( Y (s) = r (s)

10kp 1 : 200kp 0:5s + 1 Js + b = 2 10kp (kD s + 1) s + (12 + 200kp :kD )s + 20(1 + 10kp ) 1+ (0:5s + 1)(Js + b)

By choosing kp and kD any and ! n may be achieved. e. The TF to disturbance with new control: Y = W

1 20(0:5s + 1) Js + b = 2 1 10kp (kD s + 1) s + (12 + 200kp kD )s + 20(1 + 10kp ) 1+ : Js + b (0:5s + 1) ess (step in W) =

1 1 + 10kp

4033

Figure 4.40: Control system for Problem 27

As before derivative feedback a¤ects transient response only. To eliminate steady-state error we can add an integrator to the loop. This kI can be represented by replacing kp with kp + in the forward loop s and still keeping PD control in the feedback loop to obtain 20(0:5s + 1)s Y = 3 W s + (12 + 200kp kD )s2 + (20 + 200kp + 200kI kD )s + 200kI ess (step in W) = 0: 27. Consider the system shown in Fig. 4.40 with PI control. (a) Determine the transfer function from R to Y . (b) Determine the transfer function from W to Y . (c) What is the system type and error constant with respect to reference tracking? (d) What is the system type and error constant with respect to disturbance rejection? Solution: (a) Y (s) 10(kI + kp s) = : R(s) s[s(s + 1) + 20] + 10(kI + kp s) (b) Y (s) 10s = : W (s) s[s(s + 1) + 20] + 10(kI + kp s) (c) The characteristic equation is s3 + s2 + (10kp + 20)s + 10kI = 0. The Routh’s array is s3 s2 s1 s0

: 1 : 1 : 10kp + 20 10kI : 10kI

10kp + 20 10kI

For stability we must have kI > 0 and kp > kI

2.

4034 (d) System is Type 1 with respect to both r and w. The velocity constant with respect to reference tracking is Kv = kI =2 and with respect to disturbance rejection is kI :

28. Consider the second-order plant with transfer function G(s) =

1 : (s + 1)(5s + 1)

and in a unity feedback structure.

(a) Determine the system type and error constant with respect to tracking polynomial reference inputs of the system for P [D = kp ], PD kI [D = kp + kD s], and PID [D = kp + + kD s] controllers. Let s 4 kp = 19, kI = 0:5, and kD = : 19 (b) Determine the system type and error constant of the system with respect to disturbance inputs for each of the three regulators in part (a) with respect to rejecting polynomial disturbances w(t) at the input to the plant. (c) Is this system better at tracking references or rejecting disturbances? Explain your response brie‡y. (d) Verify your results for parts (a) and (b) using Matlab by plotting unit step and ramp responses for both tracking and disturbance rejection. Solution: a. This plant has no pole at the origin and DC gain of 1 so, unless the controller has such a pole, the system will be Type 0. Thus, we have: P and PD are Type 0, Kp = kp = 19;PID is Type 1, Kv = kI = 0:5 b. Again, P and PD are Type 0, Kp = kp = 19;PID is Type 1, Kv = kI = 0:5: c. Because the Types and error constants are the same, this system does the same with references as with disturbances. d. We expect the steady state error to steps to be 0 and to unit ramps to be 1=kp = 1=0:5 = 2:0: Note that steady-state is after a long time!

4035

Error res pons e to a referenc e ramp

Error to a referenc e s tep

2

0.15

1.8

1.6

0.1

1.4

1.2

0.05

1

0.8

0

0.6

0.4

-0.05

0.2

-0.1

0

10

20

30

40

50

60

70

80

90

0

100

0

10

Error to a reference step.

20

30

40

50

60

Notice that these transients are very slow. They are the consequence of a pole at s = 0:0252: A good rule of thumb is that a transient is over in 5 time constants. In this case the time constant is 1=0:0252 = 39:68: Therefore we’d expect the transient to go on for about 200 seconds! The responses to disturbances are similar:. E rror for a disturbance step

0

0

-0.2 -0.01

-0.4

-0.6

-0.02

-0.8 -0.03

-1 -0.04

-1.2

-1.4

-0.05

-1.6 -0.06

-1.8

-0.07

-2 0

10

20

30

40

50

60

70

80

90

100

Error to a disturbance step

0

10

20

30

40

50

60

70

80

90

Error to a disturbance ramp

. this case, the disturbance ramp does not excite the fast roots very much at all. 29. The DC-motor speed control shown in Fig. 4.41 is described by the differential equation y_ + 60y = 600va 1500w;

where y is the motor speed, va is the armature voltage, and w is the load torque. Assume the armature voltage is computed using the PI control law Z t va = kp e + kI edt : 0

where e = r

y:

70

80

Error to a reference ramp

100

.In

90

100

4036

Figure 4.41: D.C. Motor speed control block diagram for Problems 29 and 30

(a) Compute the transfer function from W to Y as a function of kp and kI . (b) Compute values for kp and kI so that the characteristic equation of the closed-loop system will have roots at 60 60j. Solution: (a) Transfer function: Set R = 0; then E = (s + 60)Y (s)

=

Y (s) W (s)

=

(b) For roots at

Y

kI Y (s)] 1500W (s) s 1500s 2 s + 60(1 + 10kp )s + 600kI 600[kp Y (s) +

60

j60 : comparing to the standard form: q 2 s2 + 2 ! n s + ! 2n = 0 =) s = ! n j! n 1 p ! n = 60 2 ; = 0:707 p 2 600kI = (60 2) =) kI = 12 p 60(1 + 10kp ) = 2 0:707 60 2 =) kp = 0:1

30. For the system in Problem 29, compute the following steady-state errors: (a) to a unit-step reference input; (b) to a unit-ramp reference input; (c) to a unit-step disturbance input; (d) for a unit-ramp disturbance input. (e) Verify your answers to (a) and (d) using M atlab. Note that a ramp response can be generated as a step response of a system modi…ed by an added integrator at the reference input.

4037 Solution: a. From Problem 21, kp = 0:1 and kI = 12: The DC gain of the plant is 10 so the Kv = 10kI : The system is Type 1 so the error to a step is 0. 1 1 1 b. To a unit ramp, the error is = = : Kv 10kI 120 c. For a disturbance input, the system is also Type 1. The error to a step will be 0. d. For a unit ramp disturbance input the error equals the output and is given by E

1500 W s + 60 + 600D 1500s W s2 + 60s + 600(kP s + kI ) 5 f or W = 1=s2 24

= =

ess

=

E rror response to a reference step

E rror response to a reference ramp

1.2

0.012

1 0.01

0.8 0.008

0.6 0.006 0.4

0.004 0.2

0.002

0

-0.2

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0

0.2

Error Response to a Reference Step

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

Error Response to a Reference Ramp

e. these …gures show, the error to a step goes to zero and that to a ramp goes to 1=kI = 1=120:

As

E rror response to a disturbance ramp

E rror response to a disturbance step

0

1

0

-0.05

-1

-2

-0.1 -3

-4

-0.15 -5

-6

-0.2

-7

-8

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

Error to a Disturbance Step

0.2

-0.25

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

Error to a Disturbance Ramp

0.2

4038

Figure 4.42: Satellite attitude control

And in this case, the error to a disturbance step goes to zero and the error to a disturbance ramp goes to ess = 1=kI = 0:208: 31. Consider the satellite-attitude control problem shown in Fig. 4.42 where the normalized parameters are J = 10 spacecraft inertia; N-m-sec2 =rad r = reference satellite attitude; rad. = actual satellite attitude; rad. Hy = 1 sensor scale; factor volts/rad. Hr = 1 reference sensor scale factor; volts/rad. w = disturbance torque: N-m

(a) Use proportional control, P, with D(s) = kp , and give the range of values for kp for which the system will be stable. (b) Use PD control and let D(s) = (kp + kD s) and determine the system type and error constant with respect to reference inputs. (c) Use PD control, let D(s) = (kp + kD s) and determine the system type and error constant with respect to disturbance inputs. (d) Use PI control, let D(s) = (kp + kI =s), and determine the system type and error constant with respect to reference inputs. (e) Use PI control, let D(s) = (kp + kI =s), and determine the system type and error constant with respect to disturbance inputs. (f) Use PID control, let D(s) = (kp + kI =s + kD s) and determine the system type and error constant with respect to reference inputs. (g) Use PID control, let D(s) = (kp + kI =s + kD s) and determine the system type and error constant with respect to disturbance inputs. Solution: (a) D(s) = kp ; The characteristic equation is 1 + Hy D(s)

1 =0 Js2

4039 Js2 + Hy kp = 0

r

Hy kp so that no additional damping is provided. The J system cannot be made stable with proportional control alone. or s =

j

(b) Steady-state error to reference steps. (s) r (s)

D(s) = Hr

1 Js2

1 Js2 (kp + kD s) = Hr 2 Js + (kp + kD s)Hy 1 + D(s)Hy

The parameters can be selected to make the (closed-loop) system 1 then using the FVT (assuming the system is stable. If r (s) = s stable) Hr ss = Hy and there is zero steady-state error if Hr = Hy (i.e., unity feedback). (c) Steady-state error to disturbance steps (s) 1 = W (s) Js2 + (kp + kD s)Hy 1 then using the FVT (assuming system is stable), the s 1 error is ss = . kp Hy (d) The characteristic equation is If W (s) =

1 + Hy D(s)

1 =0 Js2

With PI control, Js3 + Hy kp s + Hy kI = 0 From the Hurwitz’s test, with the s2 term missing the system will always have (at least) one pole not in the LHP. Hence, this is not a good control strategy. (e) See d above. (f) The characteristic equation with PID control is 1 + Hy (kp +

kI 1 + kD s) 2 = 0 s Js

or Js3 + Hy kD s2 + Hy kp s + Hy kI = 0

4040 There is now control over all the three poles and the system can be made stable. (s) r (s)

D(s) = Hr

1 + D(s)Hy

=

=

If

r (s)

=

1 Js2 1 Js2

kI + kD s) s kI Js2 + (k p + + kD s)Hy s Hr (kD s2 + k p s + k I ) Hr (k p +

Js3 + (kD s2 + kp s + k I )Hy

1 then using the FVT (assuming system is stable) s ss

=

Hr Hy

and there is zero steady-state error if Hr = Hy (i.e., unity feedback). In that case, the system is Type 3 and the (Jerk!) error constant is kI KJ = : J (g) The error to a disturbance is found from s (s) = 3 W (s) Js + Hy (kD s2 + k p s + k I ) 1 If W (s) = then using the FVT (assuming the system is stable), s = 0; the system is Type 1 and the error constant is Kv = Hy kp : ss

32. The unit-step response of a paper machine is shown in Fig. 4.43(a) where the input into the system is stock ‡ow onto the wire and the output is basis weight (thickness). The time delay and slope of the transient response may be determined from the …gure. (a) Find the proportional, PI, and PID-controller parameters using the Zeigler–Nichols transient-response method. (b) Using proportional feedback control, control designers have obtained a closed-loop system with the unit impulse response shown in Fig. 4.43(b). When the gain Ku = 8:556, the system is on the verge of instability. Determine the proportional-, PI-, and PID-controller parameters according to the Zeigler–Nichols ultimate sensitivity method. Solution:

4041

Figure 4.43: Paper-machine response data for problem 32

(a) From step response: L = R=

1

' 0:65 sec

d

'

0:2 = 0:33 1:25 0:65

sec

1

From Table 4.1: Controller Gain P

K=

1 RL

= 4:62

PI

K=

0:9 RL

= 4:15

P ID

K=

1:2 RL

= 5:54

L = 2:17 0:3 TI = 2L = 1:3TD = 0:5L = 0:33 TI =

(b) From the impulse response: Pu ' 2:33 sec. and from Table 4.2: Controller Gain

P PI P ID

K = 0:5Ku = 4:28 K = 0:45Ku = 3:85 K = 0:6Ku = 5:13

1 Pu = 1:86 1:2 1 1 TI = Pu = 1:12TD = Pu = 0:28 2 8

TI =

For the unity feedback system with proportional control D = kp and A process transfer function G(s) = ; s( s + 1) 33. A paper machine has the transfer function G(s) =

e 2s ; 3s + 1

where the input is stock ‡ow onto the wire and the output is basis weight or thickness.

4042

Figure 4.44: Unit impulse response for paper-machine in Problem 33

(a) Find the PID-controller parameters using the Zeigler–Nichols tuning rules. (b) The system becomes marginally stable for a proportional gain of Ku = 3:044 as shown by the unit impulse response in Fig. 4.44. Find the optimal PID-controller parameters according to the Zeigler– Nichols tuning rules. Solution: (a) From the transfer function: L = R=

d

' 2 sec

1 ' 0:33 3

sec

1

From Table 4.1: Controller Gain

P

K PI

P ID

1 1:5 RL L 0:9 K = RL = 1:35 TI = = 6:66 0:3 1:2 K = RL = 1:8 TI = 2L = 4 TD = 0:5L = 1:0 =

(b) From the impulse response: Pu ' 7 sec From Table 4.2: Controller Gain

P K

=

PI

K

=

PID

K

=

0:5Ku = 1:52 1 Pu = 5:83 1:2 1 1 0:6Ku = 1:82 TI = Pu = 3:5TD = Pu = 0:875 2 8 0:45Ku = 1:37 TI =

4043 Problems and Solutions for Section 4.4: Introduction to Digital Control 34. Compute the discrete equivalents for the following possible controllers using the trapezoid rule of Eq. (4.98). Let Ts = 0:05 in each case. (a) D1 (s) = (s + 2)=2, s+2 (b) D2 (s) = 2 , s+4 (s + 2) (c) D3 (s) = 5 , s + 10 (s + 2)(s + 0:1) (d) D4 (s) = 5 (s + 10)(s + 0:01) Solution: (a) Using the formula s

2 z 1 21z 19 we …nd D1 (z) = Ts z + 1 z+1

1:909z 1:727 z 0:8182 4:2z 3:8 (c) D3 (z) = z 0:6 4:209z 2 7:997z + 3:79 (d) D4 (z) = z 2 1:6z + 0:5997 (b) D2 (z) =

35. Give the di¤erence equations corresponding to the discrete controllers found in Problem 34 respectively (a) part 1. (b) part 2. (c) part 3. (d) part 4. Solution: (a) Reduce the z transforms to be in terms of z 1 if you want the equations in terms of past values. We have divided by the coe¢ cient of the highest power if z in the denominator to get the coe¢ cient of u(k) to U 21z 19 21 19z 1 = = and be 1:0 in each case. For part (a), E z+1 1+z 1 1 1 1 thus [1 + z ]U (z) = [21 19z ]E(z) from which, as z U (z) =) u(k 1) we get u(k) = u(k 1) + 21e(k) 19e(k 1): We have suppressed the Ts in these equations. It should properly be u(kTs ); u([k 1]Ts ); etc. (b) u(k) = 0:8182u(k (c) u(k) = 0:6u(k

1) + 1:909e(k)

1) + 4:2e(k)

1:727e(k

3:8e(k

1)

1):

4044 (d) u(k) = 1:6u(k 3:79e(k 2)

1)

0:5997u(k

2) + 4:209e(k)

7:997e(k

1) +

Chapter 5

The Root-Locus Design Method Problems and solutions for Section 5.1 1. Set up the following characteristic equations in the form suited to Evans’s root-locus method. Give L(s); a(s); and b(s) and the parameter, K; in terms of the original parameters in each case. Be sure to select K so that a(s) and b(s) are monic in each case and the degree of b(s) is not greater than that of a(s): (a) s + (1= ) = 0 versus parameter (b) s2 + cs + c + 1 = 0 versus parameter c (c) (s + c)3 + A(T s + 1) = 0 i. versus parameter A, ii. versus parameter T , iii. versus the parameter c, if possible. Say why you can or can not. Can a plot of the roots be drawn versus c for given constant values of A and T by any means at all kD s c(s) kI + ]G(s) = 0: Assume that G(s) = A where s s+1 d(s) c(s) and d(s) are monic polynomials with the degree of d(s) greater than that of c(s).

(d) 1 + [kp +

i. ii. iii. iv.

versus kp versus kI versus kD versus

5001

5002

CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD Solution:

(a) K = 1= ; a = s; b = 1

(b) K = c; a = s2 + 1; b = s + 1

(c) Part (c)

i. K = AT ; a = (s + c)3 ; b = s + 1=T ii. K = AT ; a = (s + c)3 + A; b = s iii. The parameter c enters the equation in a nonlinear way and a standard root locus does not apply. However, using a polynomial solver, the roots can be plotted versus c:

(d) Part (d)

i. K = kp A ; a = s(s + 1= )d(s) + kI (s + 1= )c(s) +

kD

s2 Ac(s);

b = s(s + 1= )c(s) ii. K = AkI ; a = s(s + 1= )d(s) + Akp s(s + 1= ) +

kD

s2 Ac(s);

b = s(s + 1= )c(s) iii. K =

AkD

; a = s(s + 1= )d(s) + Akp s(s + 1= )c(s) + AkI (s +

1= )c(s); b = s2 c(s) iv. K = 1= ; a = s2 d(s) + kp As2 c(s) + kI Asc(s); b = sd(s) + kp sAc(s) + kI Ac(s) + kD s2 Ac(s)

5003

Problems and solutions for Section 5.2

2. Roughly sketch the root loci for the pole-zero maps as shown in Fig. 5.51. Show your estimates of the center and angles of the asymptotes, a rough evaluation of arrival and departure angles for complex poles and zeros, and the loci for positive values of the parameter K. Each pole-zero map is from a characteristic equation of the form

1+K

b(s) = 0; a(s)

where the roots of the numerator b(s) are shown as small circles o and the roots of the denominator a(s) are shown as 0 s on the s-plane. Note that in Fig. 5.51(c), there are two poles at the origin and there are two poles on the imaginary axis in (f), slightly o¤ the real axis.

Solution:

5004

CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD

Root loci f or Problem 5.2 2 Imaginary Axis

Imaginary Axis

2 1 0 -1 -2 -10

1 0 -1

-5

0

-2 -6

5

-4

-2 Real Axis

Real Axis

1 0 -1 -2 -4

0.5 0 -0.5

-2

0

-1 -4

2

-2

Real Axis

2

4 Imaginary Axis

Imaginary Axis

0 Real Axis

10 5 0 -5 -10 -10

2

1 Imaginary Axis

Imaginary Axis

2

0

2 0 -2

-5

0 Real Axis

5

-4 -4

-2

0 Real Axis

2

4

5005 We had to make up some numbers to do it on Matlab, so the results partly depend on what was dreamed up, but the idea here is just get the basic rules right. (a)a(s) = s2 + s; b(s) = s + 1 Breakin(s) -3.43; Breakaway(s) -0.586 (b) a(s) = s2 + 0:2s + 1; b(s) = s + 1 Angle of departure: 135.7 Breakin(s) -4.97 (c) a(s) = s2 ; b(s) = (s + 1) Breakin(s) -2 (d) a(s) = s2 + 5s + 6; b(s) = s2 + s Breakin(s) -2.37 Breakaway(s)

-0.634

(e) a(s) = s3 + 3s2 + 4s

8

Center of asymptotes -1 Angles of asymptotes

60; 180

Angle of departure: -56.3 (f) a(s) = s3 + 3s2 + s

5; b(s) = s + 1

Center of asymptotes -.667 Angles of asymptotes

60;

180

Angle of departure: -90 Breakin(s) -2.06 Breakaway(s) 0:503 But, to get this one right, all you have to do is get the real axis segments and the 4 asymptotes going out at the right angles. You don’t know, really, where it breaks away from the real axis nor where the center of asymptotes are.

5006

CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD

3. For the characteristic equation

1+

K =0: s(s + 1)(s + 5)

(a) Draw the real-axis segments of the corresponding root locus. (b) Sketch the asymptotes of the locus for K ! 1. (c) For what value of K are the roots on the imaginary axis? (d) Verify your sketch with a MATLAB plot.

Solution:

(a) The real axis segments are 0 > (b)

=

6=3 =

2;

i

=

>

1;

5>

60; 180

(c) Ko = 30

Ro o t L o c u s

10

Ima g Axi s

5

0

-5

-1 0

-1 5

-1 0

-5

0

Re a l Ax i s

Solution for Problem 5.3

5

5007 4. Real poles and zeros. Sketch the root locus with respect to K for the equation 1 + KL(s) = 0 and the following choices for L(s). Be sure to give the asymptotes, arrival and departure angles at any complex zero or pole, and the frequency of any imaginary-axis crossing. After completing each hand sketch verify your results using MATLAB. Turn in your hand sketches and the MATLAB results on the same scales.

(a) L(s) =

(s + 2) s(s + 1)(s + 5)(s + 10)

(b) L(s) =

1 s(s + 1)(s + 5)(s + 10)

(c) L(s) =

(s + 2)(s + 6) s(s + 1)(s + 5)(s + 10)

(d) L(s) =

(s + 2)(s + 4) s(s + 1)(s + 5)(s + 10)

Solution: All the root locus plots are displayed at the end of the solution set for this problem.

(a)

=

4:67;

(b)

=

4;

i

=

45;

(c)

=

4;

i

=

90; ! o

> none

(d)

=

5;

i

=

90; ! o

> none

i

=

60;

180; ! o = 5:98

135; ! o = 1:77

5008

CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD

b

15

15

10

10 Imaginary Axis

Imaginary Axis

a

5

0

5

0

-5

-5

-10

-10

-15 -15

-10

-5 Real Axis

0

-15 -15

5

-10

15

10

10

5

0

-10

-10

-5 Real Axis

0

5

0

-5

-10

5

5

-5

-15 -15

0

d

15

Imaginary Axis

Imaginary Axis

c

-5 Real Axis

0

5

-15 -15

Root loci for Problem 5.4

-10

-5 Real Axis

5009 5. Complex poles and zeros Sketch the root locus with respect to K for the equation 1 + KL(s) = 0 and the following choices for L(s). Be sure to give the asymptotes, arrival and departure angles at any complex zero or pole, and the frequency of any imaginary-axis crossing. After completing each hand sketch verify your results using MATLAB. Turn in your hand sketches and the MATLAB results on the same scales.

(a) L(s) =

1 s2 + 3s + 10

(b) L(s) =

1 s(s2 + 3s + 10)

(c) L(s) =

(s2 + 2s + 8) s(s2 + 2s + 10)

(d) L(s) =

(s2 + 2s + 12) s(s2 + 2s + 10)

(e) L(s) =

(s2 + 1) s(s2 + 4)

(f) L(s) =

(s2 + 4) s(s2 + 1)

Solution: All the root locus plots are displayed at the end of the solution set for this problem.

(a)

=

3;

i

=

90;

(b)

=

3;

i

=

60; 180;

(c)

=

2;

i

=

180;

d

=

161:6;

(d)

=

2;

i

=

180;

d

=

18:4;

(e)

= 0;

i

=

180;

d

=

(f)

= 0;

i

=

180;

d

= 0;

d

=

90 ! o d

> none

=

180;

a

28:3 ! o = 3:16

a

a

a

=

=

=

= 0; ! o

200:7; ! o 16:8; ! o

180; ! o > none

> none > none

> none

5010

CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD

a

b 20 Imaginary Axis

Imaginary Axis

20 10 0 -10 -20 -3

-2

-1 Real Axis

0

10 0 -10 -20 -20

1

-10 0 Real Axis

c

d 4 Imaginary Axis

Imaginary Axis

4 2 0 -2 -4 -3

-2

-1 Real Axis

0

2 0 -2 -4 -3

1

-2

e

0

1

0

0.5

4 Imaginary Axis

Imaginary Axis

-1 Real Axis f

4 2 0 -2 -4 -2

10

-1

0

1

2 0 -2 -4 -1.5

Real Axis

Root loci for Problem 5.5

-1

-0.5 Real Axis

5011 6. Multiple poles at the origin Sketch the root locus with respect to K for the equation 1 + KL(s) = 0 and the following choices for L(s). Be sure to give the asymptotes, arrival and departure angles at any complex zero or pole, and the frequency of any imaginary-axis crossing. After completing each hand sketch verify your results using MATLAB. Turn in your hand sketches and the MATLAB results on the same scales.

(a) L(s) =

1 + 8)

s2 (s

(b) L(s) =

1 s3 (s + 8)

(c) L(s) =

1 s4 (s + 8)

(d) L(s) =

(s + 3) s2 (s + 8)

(e) L(s) =

(s + 3) s3 (s + 4)

(f) L(s) =

(s + 1)2 s3 (s + 4)

(s + 1)2 + 10)2 Solution:

(g) L(s) =

s3 (s

All the root locus plots are displayed at the end of the solution set for this problem.

(a)

=

2:67;

(b)

=

2;

(c)

=

1:6;

i

=

36;

(d)

=

2:5;

i

=

90; w0

(e)

=

0:33;

(f)

=

3;

i

=

90; w0 =

(g)

=

6;

i

=

60; 180; w0 =

i

i

=

=

i

60;

45;

=

180; w0

135; w0

60;

> none

> none

108; w0

> none

> none 180; w0

> none

1:414 1:31; 7:63

5012

CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD 10

Imag Axis

a 0

-10

-10

-5 Real Axis c

-10

-5 Real Axis

Imag Axis

10

-5 Real Axis

Imag Axis

20

0

-8

-5 Real Axis

-4 -2 Real Axis

0

2

-6

-4 -2 Real Axis

0

2

f

-20

-8

0

-10

-6

0

g

-20

0

d

20

0

-10

-5 Real Axis

0

-20

0

e

-10

-10

20

0

-10

b

0

-10

0

Imag Axis

Imag Axis

10

Imag Axis

Imag Axis

10

0

Solution for Problem 5.6

5013 7. Mixed real and complex poles Sketch the root locus with respect to K for the equation 1 + KL(s) = 0 and the following choices for L(s). Be sure to give the asymptotes, arrival and departure angles at any complex zero or pole, and the frequency of any imaginary-axis crossing. After completing each hand sketch verify your results using MATLAB. Turn in your hand sketches and the MATLAB results on the same scales.

(a) L(s) =

(b) L(s) =

(s + 2) s(s + 10)(s2 + 2s + 2)

s2 (s

(s + 2) + 10)(s2 + 6s + 25)

(c) L(s) =

(s + 2)2 s2 (s + 10)(s2 + 6s + 25)

(d) L(s) =

(s + 2)(s2 + 4s + 68) s2 (s + 10)(s2 + 4s + 85)

(e) L(s) =

[(s + 1)2 + 1] + 2)(s + 3)

s2 (s

Solution: All the plots are attached at the end of the solution set.

(a)

=

3:33;

(b)

=

3:5;

(c)

=

4;

i

=

60;

(d)

=

4;

i

=

90; w0

(e)

=

1:5;

i

i

i

=

=

=

60; 45;

180; w0 = 135; w0

180; w0 =

90; w0

2:32;

d

=

6:34

> none;

d

=

103:5

6:41;

> none; > none;

d

d

= a

=

14:6

106; =

71:6

a

=

253:4

5014

CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD 20

10

10 plot b

plot a

5

Imag Axis

0

0

-10

-20

plot c

5

Imag Axis

Imag Axis

10

-5

-10

-5 Real Axis

0

-10

40

-5

-10

-5 Real Axis

-10

0

5 plot e

plot d

Imag Axis

Imag Axis

20

0

0

-20

-40

-10

-5 0 Real Axis

-5 -4

0

-2 0 Real Axis

2

Solution for Problem 5.7

-10

-5 Real Axis

0

5015 8. Right half plane poles and zeros Sketch the root locus with respect to K for the equation 1 + KL(s) = 0 and the following choices for L(s). Be sure to give the asymptotes, arrival and departure angles at any complex zero or pole, and the frequency of any imaginary-axis crossing. After completing each hand sketch verify your results using MATLAB. Turn in your hand sketches and the MATLAB results on the same scales.

1 s+2 ; The model for a case of magnetic levitation 2 s + 10 s 1 with lead compensation.

(a) L(s) =

1 s+2 ; The magnetic levitation system with ins(s + 10) (s2 1) tegral control and lead compensation.

(b) L(s) =

(c) L(s) =

s

1 s2

s2 + 2s + 1 : What is the largest value that can s(s + 20)2 (s2 2s + 2) be obtained for the damping ratio of the stable complex roots on this locus?

(d) L(s) =

(e) L(s) =

(f) L(s) =

s(s

(s + 2) ; 1)(s + 6)2

(s

1 1)[(s + 2)2 + 3]

Solution:

(a)

=

4;

i

=

90; w0

(b)

=

4;

i

=

60; 180; w0

(c)

=

1;

i

=

180; w0

(d)

=

12;

(e)

=

3;

i

=

60; 180; w0

(f)

=

1;

i

=

60; 180; w0 =

i

=

> none > none

> none

60; 180; w0 =

3:24; 15:37;

d

=

> none 1:732;

d

=

40:9

92:4

5016

CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD 10

10

1

plot a

0

-5

0.5 Imag Axis

5 Imag Axis

Imag Axis

5

-10

plot c

plot b

0

-5

-10

-10

-5 0 Real Axis

plot d

0

-0.5

-10

-5 Real Axis

-1 -1

0

10

10

plot f

Imag Axis

0 -5 -10

2 Imag Axis

5

5

2

4 plot e

15

Imag Axis

0 1 Real Axis

0

-5

0

-2

-15 -20 -5

0 Real Axis

5

-10 -8

-6

-4 -2 0 Real Axis

Solution for Problem 5.8

2

-4

-2

0 Real Axis

2

5017 9. Put the characteristic equation of the system shown in Fig. 5.52 in root locus form with respect to the parameter and identify the corresponding L(s); a(s); and b(s): Sketch the root locus with respect to the parameter , estimate the closed-loop pole locations and sketch the corresponding step responses when = 0; 0:5, and 2. Use MATLAB to check the accuracy of your approximate step responses.

Figure 5.52: Control system for problem 9

Solution: The characteristic equation is s2 +2s+5+5 s = 0 and L(s) = the root locus and step responses are plotted below.

root locus

s : s2 + 2s + 5

StepResponse 1.4

2

1.5

alpha=0

1.2

1

1

alpha=2 Amplitude

Imag Axis

0.5

0

0.8

alpha=0.5

0.6 -0.5 0.4

-1

-1.5

0.2

-2 -4

-3

-2 Real Axis

-1

0

0

0

1

2

3 4 Time(sec)

5

6

7

5018

CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD

10. Use the MATLAB function rltool to study the behavior of the root locus of 1 + KL(s) for

L(s) =

(s + a) s(s + 1)(s2 + 8s + 52)

as the parameter a is varied from 0 to 10, paying particular attention to the region between 2:5 and 3:5. Verify that a multiple root occurs at a complex value of s for some value of a in this range.

Solution:

For small values of ; the locus branch from 0; 1 makes a circular path around the zero and the branches from the complex roots curve o¤ toward the asymptotes. For large values of the branches from the complex roots break into the real axis and those from 0; 1 curve o¤ toward the asymptotes. At about = 3:11 these loci touch corresponding to complex multiple roots.

5019 11. Use the Routh criterion to …nd the range of the gain K for which the systems in Fig. 5.53 are stable and use the root locus to con…rm your calculations.

Figure 5.53: Feedback systems for problem 11

Solution: (a) The system is stable for 0 K 478:226 The root locus of the system and the location of the roots at the crossover points are shown in the plots (b) There is a pole in the right hand plane thus the system is unstable for all values of K as shown in the last plot. Root Locus

Root Locus

6 10 I ma g A xi s

I ma g A xi s

4 2 0 -2

5 0 -5

-4 -10 -6 -15

-10

-5 Real Axis

0

5

Solution for Problem 11

-10

-5 Real Axis

0

5

5020

CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD

12. Sketch the root locus for the characteristic equation of the system for which L(s) =

(s + 2) ; s(s + 1)(s + 5)

and determine the value of the root-locus gain for which the complex conjugate poles have a damping ratio of 0.5. Solution: Plot the system on Matlab using rlocus(sys), and use [K]= rloc…nd(sys) to pick the gain where the damping ratio = 0.5. Find that K = 14 (approximately).

Root Locus 15 0.38

0.26

0.19

0.13

0.085

0.04

14

12

10

10 0.52 8

6 5

Imaginary A xis

0.8

4

2

0

2

4

0.8 -5

6

8 -10

0.52 10

12

0.26

0.38 -15 -6

-5

-4

0.19 -3

0.13 -2

0.085 -1

0.04

14 0

Real Axis

Root locus with 0.5 damping marked

1

5021 13. For the system in Fig. 5.54:

Figure 5.54: Feedback system for problem 13

(a) Find the locus of closed-loop roots with respect to K. (b) Is there a value of K that will cause all roots to have a damping ratio greater than 0:5? (c) Find the values of K that yield closed-loop poles with the damping ratio = 0:707. (d) Use MATLAB to plot the response of the resulting design to a reference step. Solution: (a) The locus is plotted below (b) There is a K which will make the ’dominant’ poles have damping 0.5 but none that will make the poles from the resonance have that much damping. (c) Using rloc…nd, the gain is about 35. (d) The step response shows the basic form of a well damped response with the vibration of the response element added. Root Locus 0.84

8 0.92 6

0.74

0.6

1.2 0.965

4

1 0.99

Am pli tude

Im ag Ax is

StepResponse

1.4

0.42 0.22

2 200 17.5

15 12.5

10

7.5

5

2.5

-2

0.8 0.6

0.99 0.4

-4 -6 -8 -20

0.965 0.2 0.92

0.84 -15

0.74 0.6 0.42 0.22 -10 -5 Real Axis

0

0

0

1

2

3 Time(sec)

Root locus and step response for Problem 5.13

4

5

6

5022

CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD

14. For the feedback system shown in Fig. 5.55, …nd the value of the gain K that results in dominant closed-loop poles with a damping ratio = 0:5.

Figure 5.55: Feedback system for Problem 14

Solution:

Use block diagram reduction to …nd the characteristic equation of the closed loop system, then divide that up into terms with and without K to 10s …nd the root locus form, where L(s) = 2 : Plugging into Matlab s + s + 10 and using rloc…nd produces the required gain to be K = 0:22:The locus is

5023

Root Locus 4 0.84

0.72

0.58

0.44

0.3

0.14

3 0.92

2

0.98

Imaginary Axis

1

6 0

5

4

3

2

1

-1 0.98

-2

0.92 -3

0.72

0.84 -4 -6

-5

-4

0.58 -3

0.44 -2

0.3 -1

0.14 0

Real A xis

Root locus with 0.5 damping marked

1

5024

CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD

Problems and solutions for Section 5.3 15. A simpli…ed model of the longitudinal motion of a certain helicopter near hover has the transfer function G(s) =

9:8(s2 0:5s + 6:3) : (s + 0:66)(s2 0:24s + 0:15)

and the characteristic equation 1 + D(s)G(s) = 0. Let D(s) = kp at …rst. (a) Compute the departure and arrival angles at the complex poles and zeros. (b) Sketch the root locus for this system for parameter K = 9:8kp :Use axes -4 x 4. 3 y 3; (c) Verify your answer using MATLAB. Use the command axis([-4 4 -3 3]) to get the right scales. (d) Suggest a practical (at least as many poles as zeros) alternative compensation D(s) which will at least result in a stable system. Solution: (a)

= :92;

= 180; ' = 63:83;

=

26:11

Root Locus 3

2

D = 9.8k p D ouble lead

Imag Axis

1

0

-1

-2

(b)

-3 -4

-3

-2

-1

0

1

2

3

4

Real A xis

Problem 5.15(b) (c) For this problem a double lead is needed to bring the roots into the left half-plane. The plot shows the rootlocus for control for. Let (s + :66)(s + :33) D= : (s + 5)2

5025 Root Locus 3

2

Imag Axis

1

0

-1

-2

(d)

-3 -4

-3

-2

-1

0

1

Real A xis

Problem 5.15(d)

2

3

4

5026

CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD

Figure 5.56: Control system for problem 5.16

16. For the system given in Fig. 5.56,

(a) plot the root locus of the characteristic equation as the parameter K1 is varied from 0 to 1 with = 2. Give the corresponding L(s); a(s); and b(s): (b) Repeat part (a) with value?

= 5. Is there anything special about this

(c) Repeat part (a) for …xed K1 = 2 with the parameter K = from 0 to 1. Solution: The root locus for each part is attached at the end.

(a) L(s) =

0:75 S(0:1S 2 +1:1S+1:8)

=

(b) L(s)= S(0:1S 20:75 +1:4S+4:5) = S(0:1S+0:9) (c) L(s)= 0:1S^3+0:9S+1:5 =

a(s) b(s)

a(s) b(s)

a(s) b(s)

varying

5027 Root Locus

Root Locus 20

20

plot b

plot a 10 Imag Axis

Imag Axis

10 0 -10

0

-10

-20 -20 -30

-20

-10

0

10

-25

Real Axis

plot c

Imag Axis

0.5 0 -0.5 -1 -15

-10

-15

-10 Real Axis

Root Locus

1

-20

-5

0

Real Axis

Solution for problem 5.16

-5

0

5

5028

CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD

Figure 5.57: Control system for problem 17

17. For the system shown in Fig. 5.57, determine the characteristic equation and sketch the root locus of it with respect to positive values of the parameter c. Give L(s), a(s); and b(s) and be sure to show with arrows the direction in which c increases on the locus.

(a) Solution: s2 + 9 a(s) = 3 s + 144s b(s)

Root Locus

10

5

Im ag Axis

L(s) =

0

-5

-10

-18

-16

-14

-12

-10

-8

-6

-4

Real Axis

Solution for problem 5.17

-2

0

5029 18. Suppose you are given a system with the transfer function

L(s) =

(s + z) ; (s + p)2

where z and p are real and z > p. Show that the root-locus for 1+KL(s) = 0 with respect to K is a circle centered at z with radius given by

r = (z

p)

Hint. Assume s + z = rej and show that L(s) is real and negative for real under this assumption.

Solution:

s + z = (z

G=

((z

p)ej

(z p)ej p)ej + p

= z)2 (z

(z p)ej = p)2 (ej 1)2 (z

1 p)( 4)( e

j =2

e 2j

j =2

)2

1 1 Because z > p; this function is real and negative 4(z p) (sin( =2))2 for real and therefore these points are on the locus. =

5030

CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD

19. The loop transmission of a system has two poles at s = 1 and a zero at s = 2. There is a third real-axis pole p located somewhere to the left of the zero. Several di¤erent root loci are possible, depending on the exact location of the third pole. The extreme cases occur when the pole is located at in…nity or when it is located at s = 2. Give values for p and sketch the three distinct types of loci.

Root Locus

Root Locus 0.5

20

pole is at negative infinity

pole is at s=-2 Imag Axis

Imag Axis

10 0

0

-10 -20 -0.5 -100

-80

-60

-40

-20

0

-2

Real Axis

Root Locus

Imag Axis

pole is slightly to the left of zero

0

-5 -2.5

-2

-1.5

-1 Real Axis

5

-3

-1.5

-1

-0.5

0

Real Axis

Solution for problem 5.19

-0.5

0

5031 20. For the feedback con…guration of Fig. 5.58, use asymptotes, center of asymptotes, angles of departure and arrival, and the Routh array to sketch root loci for the characteristic equations of the following feedback control systems versus the parameter K: Use MATLAB to verify your results.

(a) G(s) =

1 s(s + 1 + 3j)(s + 1

(b) G(s) =

1 ; s2

(c) G(s) =

(s + 5) ; (s + 1)

(d) G(s) =

(s + 3 + 4j)(s + 3 4j) ; s(s + 1 + 2j)(s + 1 2j)

3j)

;

H(s) =

H(s) =

H(s) =

s+7 s+3

H(s) = 1 + 3s

Figure 5.58: Feedback system for problem 20

Solution:

s+1 s+3

s+2 s+8

5032

CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD Root Locus

Root Locus 5

plot a

plot b

Imag Axis

Imag Axis

5

0

0

-5 -5 -10

-5 Real Axis

0

-3

-2.5

-2

Root Locus

-1.5 Real Axis

-1

-0.5

0

-1

-0.5

0

Root Locus 4

1.5

plot c

plot d 2

0.5

Imag Axis

Imag Axis

1

0 -0.5

0

-2

-1 -1.5

-4 -6

-4 Real Axis

-2

0

-3

Solution for problem 5.20

-2.5

-2

-1.5 Real Axis

5033 21. Consider the system in Fig. 5.59.

Figure 5.59: Feedback system for problem 5.21

(a) Using Routh’s stability criterion, determine all values of K for which the system is stable. (b) Use Matlab to …nd the root locus versus K. Find the values for K at imaginary-axis crossings. Solution: (a) a. (b)

d

0

K

40

=

161:6

a

=0

At imaginary axis crossing

s= j1:8186

k = 6:2758

Root locus is attached for reference. Root Locus

6

4

Imag Axis

2

0

-2

-4

-6

-7

-6

-5

-4

-3

-2

-1

0

1

Real A xis

Root locus for problem 21

2

3

5034

CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD

Problems and solutions for Section 5.4

22. Let

G(s) =

1 (s + 2)(s + 3)

and D(s) = K

s+a : s+b

Using root-locus techniques, …nd values for the parameters a; b, and K of the compensation D(s) that will produce closed-loop poles at s = 1 j for the system shown in Fig. 5.60.

Figure 5.60: Unity feedback system for Problems 5.22 to 5.28 and 5.33

Solution:

Since the desired poles are slower than he plant, we will use PI control. The solution is to cancel the pole at -3 with the zero and set the gain to K = 2: Thus, p = 0; z = 3; K = 2:

5035 23. Suppose that in Fig. 5.60,

G(s) =

1 s(s2 + 2s + 2)

and D(s) =

K : s+2

Sketch the root-locus with respect to K of the characteristic equation for the closed-loop system, paying particular attention to points that generate multiple roots if KL(s) = D(s)G(s).

Solution:

The locus is plotted below. The roots all come together at s = K = 1:

Root Locus 1.5

1

Ima g Axi s

0.5

0

-0.5

-1

-1.5

-3

-2.5

-2

-1.5

-1 Real Axis

-0.5

0

Root locus for Problem 23

0.5

1

1 at

5036

CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD

24. Suppose the unity feedback system of Fig. 5.60 has an open-loop plant s+z given by G(s) = 1=s2 . Design a lead compensation D(s) = K to be s+p added in series with the plant so that the dominant poles of the closed-loop system are located at s = 2 2j.

Solution:

Setting the pole of the lead to be at p = 20; the zero is at z = with a gain of K = 72: The locus is plotted below.

Root Locus 3

0.92

0.84

0.74

0.6

0.42

0.22

2 0.965

Imag Axis

1

0.99

7 0

-1

6

5

-6

-5

4

3

2

1

0.99

0.965 -2

-3 -7

0.84

0.92

0.74 -4

-3

0.6 -2

0.42

0.22 -1

Real Axis

Root locus for Problem 24

0

1

1:78

5037 25. Assume that the unity feedback system of Fig. 5.60 has the open-loop plant

G(s) =

1 : s(s + 3)(s + 6)

Design a lag compensation to meet the following speci…cations:

The step response settling time is to be less than 5 sec. The step response overshoot is to be less than 17%. The steady-state error to a unit ramp input must not exceed 10%. Solution: The overshoot speci…cation requires that damping be 0:5 and the settling time requires that ! n > 1:8: From the root locus plotted below, these can be met at K = 28 where the ! n = 2: With this gain, the Kv = 28=18 = 1:56: To get a Kv = 10; we need a lag gain of about 6:5: Selecting the lag zero to be at 0:1 requires the pole to be at 0:1=6:5 = 0:015: To meet the overshoot speci…cations, it is necessary to select a smaller K and set p = 0:01: Other choices are of course possible. The step response of this design is plotted below.

6

0.84

0.74

1.4

Root Locus 0.6 0.42 0.22

0.92

1.2

4

Ima g Axi s

0.965 2 102

-2

1

0.99

0.8

10

8

6

4

2

0.6 0.99

0.4 0.965

-4

0.2 0.92

-6

0.84 -10

0.74

0.6 -5 Real Axis

0.42 0.22 0

00

5

10

15

Root locus and step response for Problem 5.25

20

25

5038

CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD

26. A numerically controlled machine tool positioning servomechanism has a normalized and scaled transfer function given by

G(s) =

1 : s(s + 1)

Performance speci…cations of the system in the unity feedback con…guration of Fig. p 5.60 are satis…ed if the closed-loop poles are located at s = 1 j 3.

(a) Show that this speci…cation cannot be achieved by choosing proportional control alone, D(s) = kp .

(b) Design a lead compensator D(s) = K

s+z that will meet the specis+p

…cation. Solution:

(a) With proportional control, the poles have real part at s =

(b) To design a lead, we select the pole to be at p = zero and gain to be z = 3; k = 12:

:5:

10 and …nd the

5039 27. A servomechanism position control has the plant transfer function G(s) =

10 : s(s + 1)(s + 10)

You are to design a series compensation transfer function D(s) in the unity feedback con…guration to meet the following closed-loop speci…cations: The response to a reference step input is to have no more than 16% overshoot. The response to a reference step input is to have a rise time of no more than 0.4 sec. The steady-state error to a unit ramp at the reference input must be less than 0.02 (a) Design a lead compensation that will cause the system to meet the dynamic response speci…cations. (b) If D(s) is proportional control, D(s) = kp ; what is the velocity constant Kv ? (c) Design a lag compensation to be used in series with the lead you have designed to cause the system to meet the steady-state error speci…cation. (d) Give the MATLAB plot of the root locus of your …nal design. (e) Give the MATLAB response of your …nal design to a reference step . Solution: (a) Setting the lead pole at p = 60 and the zero at z = 1; the dynamic speci…cations are met with a gain of 245 resulting in a Kv = 4: (b) Proportional control will not meet the dynamic spec. The Kv of the lead is given above. (c) To meet the steady-state requirement, we need a new Kv = 50; which is an increase of 12:5: If we set the lag zero at z = :4; the pole needs to be at p = 0:032: (d) The root locus is plotted below. (e) The step response is plotted below.

5040

CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD Lead root locus 4 0.92

0.84

0.74

Lead Step response 1.5

0.6 0.420.22

0.965 2 1 8

6

4

Amplitude

Imag Axis

0.99 10 0

2

0.99

0.5

-2 0.965 -4

0.84

0.92

-10

-8

0.74

-6

0.6 0.420.22

-4

-2

0 0

2

0

0.2

0.4

Real Axis

Lead-lag root locus 4 0.92

0.84

0.74

0.6

0.8

1

1.2

Time (sec)

Lead-lag step response 1.5

0.6 0.420.22

0.965 2 1 8

6

4

Amplitude

Imag Axis

0.99 10 0

2

0.99

0.5

-2 0.965 -4 -10

0.84

0.92 -8

-6

0.74

0.6 0.420.22

-4

-2

0 0

2

0

Real Axis

1

2

3

4

Time (sec)

Solution to Problem 27

5

6

7

5041 28. Assume the closed-loop system of Fig. 5.60 has a feed forward transfer function G(s) given by

G(s) =

1 : s(s + 2)

Design a lag compensation so that the dominant poles of the closed-loop system are located at s = 1 j and the steady-state error to a unit ramp input is less than 0.2.

Solution:

The poles can be put in the desired location with proportional control alone, with a gain of kp = 2 resulting in a Kv = 1: To get a Kv = 5; we s + 0:1 add a compensation with zero at 0:1 and a pole at 0:02: D(s) = 2 : s + 0:02

5042

CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD

29. An elementary magnetic suspension scheme is depicted in Fig. 5.61. For small motions near the reference position, the voltage e on the photo detector is related to the ball displacement x (in meters) by e = 100x. The upward force (in newtons) on the ball caused by the current i (in amperes) may be approximated by f = 0:5i + 20x. The mass of the ball is 20 g, and the gravitational force is 9.8 N/kg. The power ampli…er is a voltage-to-current device with an output (in amperes) of i = u + V0 .

Figure 5.61: Elementary magnetic suspension

(a) Write the equations of motion for this setup. (b) Give the value of the bias V0 that results in the ball being in equilibrium at x = 0. (c) What is the transfer function from u to e? (d) Suppose the control input u is given by u = Ke. Sketch the root locus of the closed-loop system as a function of K. U (e) Assume that a lead compensation is available in the form = E s+z D(s) = K : Give values of K; z; and p that yields improved s+p performance over the one proposed in part (d). Solution: (a) m• x = 20x + 0:5i Vo ) 0:196:

mg: Substituting numbers, 0:02• x = 20x + 0:5(u +

(b) To have the bias cancel gravity, the last two terms must add to zero. Thus Vo = 0:392:

5043 (c) Taking transforms of the equation and substituting e = 100x;

E 2500 = 2 U s 1000

(d) The locus starts at the two poles symmetric to the imaginary axis, meet at the origin and cover the imaginary axis. The locus is plotted below.

(e) The lead can be used to cancel the left-hand-plane zero and the pole at m 150 which will bring the locus into the left-hand plane where K can be selected to give a damping of, for example 0.7. See the plot below.

Root l oc i for Problem 5.31 60

Im ag Ax i s

0.92

40

0.965

20

0.99

0

140

-20

0.99

-40

0.965

0.84

120

100

0.92 -60 -150

0.74

80

0.84

0.6

60

40

0.74

-100

0.42

0.6

0.22

20

0.42

-50 Real Ax is

Root loci for Problem 29

0.22 0

5044

CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD n

30. A certain plant with the non minimum phase transfer function G(s) =

4 2s ; +s+9

s2

is in a unity positive feedback system with the controller transfer function D(s):

(a) Use MATLAB to determine a (negative) value for D(s) = K so that the closed-loop system with negative feedback has a damping ratio = 0:707. (b) Use MATLAB to plot the system’s response to a reference step. Solution:

(a) With all the negatives, the problem statement might be confusing. With the G(s) as given, MATLAB needs to plot the negative locus, which is the regular positive locus for G: The locus is plotted below. The value of gain for closed loop roots at damping of 0:7 is k = 1:04 (b) The …nal value of the step response plotted below is 0:887. To get a positive output we would use a positive gain in positive feedback.

Root Locus 4

0.84

0.92

0.74

0.6

StepResponse 0.4

0.42 0.22

0.2

3 0.965 2

0 0.99

100

8

6

4

Amplitude

Imag Axis

1 2

-0.2

-0.4 -1 0.99 -0.6

-2 0.965 -3 -4 -10

-0.8 0.92

0.84 -8

0.74 -6

0.6 -4 Real Axis

0.42 0.22 -2

0

2

-1

0

Solutions for Problem 30

1

2 Time(sec)

3

4

5045 31. Consider the rocket-positioning system shown in Fig. 5.62.

Figure 5.62: Block diagram for rocket-positioning control system

(a) Show that if the sensor that measures x has a unity transfer function, the lead compensator s+2 H(s) = K s+4 stabilizes the system. (b) Assume that the sensor transfer function is modeled by a single pole with a 0:1 sec time constant and unity DC gain. Using the root-locus procedure, …nd a value for the gain K that will provide the maximum damping ratio. Solution: (a) The root locus is plotted below and lies entirely in the left-half plane. However the maximum damping is 0:2: (b) At maximum damping, the gain is K = 6:25 but the damping of the complex poles is only 0:073: A practical design would require much more lead. Root Locus 0.19 0.13

0.26 10

3

0.09 0.06 0.03 12 10

0.4

Im a g Ax i s

2 0

Im a g Ax i s

6 4

4

0.65

8

0.4

-4

0.19 -3

-2 Real Axis

0.13

0.09 0.06 0.03 12 -1 0

0.88

1.5

2

1

0.5

0.97 -1

1 1.5

0.88 -2

2 0.76

-3-3

Loci for problem 31 32. For the system in Fig. 5.63:

0.5

0

10 0.26

3 2.5

0.97

6 -10

Root Locus 0.36 0.24 0.12

1

2 -5

0.48

2

8

5 0.65

0.62 0.76

0.62

0.48 -2

0.36 0.24 0.12 -1 Real Axis

2.5 30

1

CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD

Figure 5.63: Control system for Problem 32

(a) Sketch the locus of closed-loop roots with respect to K. (b) Find the maximum value of K for which the system is stable. Assume K = 2 for the remaining parts of this problem. (c) What is the steady-state error (e = r

y) for a step change in r?

(d) What is the steady-state error in y for a constant disturbance w1 ? (e) What is the steady-state error in y for a constant disturbance w2 ? (f) If you wished to have more damping, what changes would you make to the system? Solution: (a) For the locus, L(s) =

100(s + 1) : The locus is plotted below. s2 (s2 + 12s + 40) Root Locus

6

0.92

0.84

0.74

0.6

0.42

0.22

4 0.965

2

Imag Axis

5046

0.99

14 0

-2

-4

-6 -14

12

10

-12

-10

8

6

4

2

0.99

0.965

0.84

0.92

0.74 -8

-6

0.6 -4

0.42

0.22 -2

0

Real A xis

Locus for Problem 32 (b) The maximum value of K for stability is K = 3:35:

2

5047 200 : s2 (s2 + 12 + 40) + 200s Thus the system is type 1 with Kv = 1: If the velocity feedback 200 were zero, the system would be type 2 with Ka = = 5: 40 100s2 Y = 2 2 : The (d) The transfer function W1 s (s + 12s + 40) + 200(s + 1) system is thus type 2 with Ka = 100: Y 100 (e) The transfer function = 2 2 : The W2 s (s + 12s + 40) + 200(s + 1) system here is type 0 with Kp = 1: (c) The equivalent plant with unity feedback is G0 =

(f) To get more damping in the closed-loop response, the controller needs to have a lead compensation. 33. Consider the plant transfer function G(s) =

s2 [mM s2

bs + k + (M + m)bs + (M + m)k]

to be put in the unity feedback loop of Fig. 5.60. This is the transfer function relating the input force u(t) and the position y(t) of mass M in the non-collocated sensor and actuator problem. In this problem we will use root-locus techniques to design a controller D(s) so that the closedloop step response has a rise time of less than 0.1 sec and an overshoot of less than 10%. You may use MATLAB for any of the following questions. (a) Approximate G(s) by assuming that m = 0, and let M = 1, k = 1, b = 0:1, and D(s) = K. Can K be chosen to satisfy the performance speci…cations? Why or why not? (b) Repeat part (a) assuming D(s) = K(s + z), and show that K and z can be chosen to meet the speci…cations. (c) Repeat part (b) but with a practical controller given by the transfer function p(s + z) D(s) = K ; s+p and pick p so that the values for K and z computed in part (b) remain more or less valid. (d) Now suppose that the small mass m is not negligible, but is given by m = M=10. Check to see if the controller you designed in part (c) still meets the given speci…cations. If not, adjust the controller parameters so that the speci…cations are met. Solution: (a) The locus in this case is the imaginary axis and cannot meet the specs for any K:

5048

CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD (b) The specs require that > 0:6; ! n > 18: Select z = 15 for a start. The locus will be a circle with radius 15: Because of the zero, the overshoot will be increased and Figure 3.32 indicates that we’d better make the damping greater than 0.7. As a matter of fact, experimentation shows that we can lower the overshoot of less than 10% only by setting the zero at a low value and putting the poles on the real axis. The plot shows the result if D = 25(s + 4): s+4 : (c) In this case, we take D(s) = 20 :01s + 1 (d) With the resonance present, the only chance we have is to introduce a notch as well as a lead. The compensation resulting in the plots s + 4 s2 =9:25 + s=9:25 + 1 : The design gain shown is D(s) = 11 (:01s + 1) s2 =3600 + s=30 + 1 was obtained by a cycle of repeated loci, root location …nding, and step responses. Refer to the …le ch5p35.m for the design aid. 1.5

Root Locus

Imag Axis

20

1

10 0

0.5

-10 -20 -60

-50

-40

-30

-20

-10

0

0

0

0.2

0.4

0.6

0.8

0

0.2

0.4

0.6

0.8

Real Axis

1.5

Root Locus

Imag Axis

20

1

10 0

0.5

-10 -20 -60

-50

-40

-30

-20

-10

0

0

Real Axis Root Locus

1

Imag Axis

4 2

0.5

0 -2 -4 -10

-8

-6

-4

-2

0

2

0

0

0.2

0.4

0.6

0.8

1

Real Axis

Root loci and step responses for Problem 33 34. Consider the type 1 system drawn in Fig. 5.64. We would like to design the compensation D(s) to meet the following requirements: (1) The steadystate value of y due to a constant unit disturbance w should be less than 4 = 0:7. Using root-locus techniques: 5 , and (2) the damping ratio (a) Show that proportional control alone is not adequate.

5049

Figure 5.64: Control system for problem 34

(b) Show that proportional-derivative control will work. (c) Find values of the gains kp and kD for D(s) = kp + kD s that meet the design speci…cations. Solution: (a) To meet the error requirements, the input to D(s) is -0:8 and the output must be 1:0 to cacel the disturbance. Thus the controller dc gain must be at least 1:25: With proportional control and a closed loop damping of 0:70, the gain is 0:5 which is too low. (b) With PD control, the characteristic equation is s2 + (1 + kD )s + kp : Setting kp = 1:25 and damping 0:7; we get kD = 0:57: The root loci and disturbance step response are plotted below. (c) The gains are kp = 1:25; kD = 0:57: Root Locus

1.5

Stepresponseforproblem5.36

0.9 0.8

1 0.7 0.6 Amplitude

Imag Ax is

0.5

0

-0.5

0.5 0.4 0.3 0.2

-1 0.1 -1.5

-3

-2

-1 Real Axis

0

1

0

0

2

4 Time(sec)

Solution for problem 34

Problems and solutions for Section 5.5

6

8

5050

CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD

35. Consider the positioning servomechanism system shown in Fig. 5.65, where ei = Kpot i ; eo = Kpot o ; Kpot = 10V=rad; T = motor torque = Kt ia ; Kt = torque constant = 0:1 N m=A; = Ke Ra = armature resistance = 10 ; Gear ratio = 1 : 1; JL + Jm = total inertia = 10 C = 200 F; va = KA (ei ef ):

3

kg m2 ;

Figure 5.65: Positioning servomechanism

(a) What is the range of the ampli…er gain KA for which the system is stable? Estimate the upper limit graphically using a root-locus plot. (b) Choose a gain KA that gives roots at = 0:7. Where are all three closed-loop root locations for this value of KA ? Solution:

(a) 0 < K < 110

5051 Root locus for problem 5.37 3

2

Im ag Axis

1

0

-1

-2

-3 -7

-6

-5

-4

-3

-2

-1

0

1

Real Axis

Root locus for problem 35 K = 10:; poles are at s =

10:05;

0:475

j0:475:

36. We wish to design a velocity control for a tape-drive servomechanism. The transfer function from current I(s) to tape velocity (s) (in millimeters per millisecond per ampere) is (s) 15(s2 + 0:9s + 0:8) = : I(s) (s + 1)(s2 + 1:1s + 1) We wish to design a type 1 feedback system so that the response to a reference step satis…es tr

4msec;

ts

15msec;

Mp

0:05

(a) Use the integral compensator kI =s to achieve type 1 behavior, and sketch the root-locus with respect to kI . Show on the same plot the region of acceptable pole locations corresponding to the speci…cations. (b) Assume a proportional-integral compensator of the form kp (s + )=s, and select the best possible values of kp and you can …nd. Sketch the root-locus plot of your design, giving values for kp and , and the velocity constant Kv your design achieves. On your plot, indicate the closed-loop poles with a dot , and include the boundary of the region of acceptable root locations. Solution: (a) The root locus is plotted with the step response below in the …rst row. (b) The zero was put at s = 1:7 and the locus and step response are plotted in the second row below

5052

CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD

Root locus

Integral control

1

0.5

Amplitude

Imag Axis

0.8 0

0.6 0.4

-0.5 0.2 0 -2

-1.5

-1

-0.5

0

0

2

4

6

Real Axis

Time (sec)

Root Locus

PI control

8

10

8

10

1 0.5

Amplitude

Imag Axis

0.8 0

0.6 0.4

-0.5 0.2 0 -2

-1.5

-1

-0.5

0

0

Real Axis

2

4

6

Time (sec)

Solution for problem 36 37. The normalized, scaled equations of a cart as drawn in Fig. 5.66 of mass mc holding an inverted uniform pendulum of mass mp and length ` with no friction are • = v (5.1) y• + =v 3mp is a mass ratio bounded by 0 < < 0:75. Time is 4(mc + mp ) 3g(mc + mp ) measured in terms of = ! o t where ! 2o = : The cart motion, `(4mc + mp ) 3x y; is measured in units of pendulum length as y = and the input is 4` u force normalized by the system weight, v = : These equations g(mc + mp ) can be used to compute the transfer functions where

=

V

=

1 s2

1

Y s2 1 + = 2 2 V s (s 1)

(5.2)

(5.3)

In this problem you are to design a control for this system by …rst closing a loop around the pendulum, Eq.(5.2) and then, with this loop closed,

5053

Figure 5.66: Figure of cart-pendulum for Problem 37

closing a second loop around the cart plus pendulum Eq.(5.3). For this problem, let the mass ratio be mc = 5mp : (a) Draw a block diagram for the system with V input and both Y and as outputs. s+z for the loop to (b) Design a lead compensation Dp (s) = Kp s+p cancel the pole at s = 1 and place the two remaining poles at 4 j4: The new control is U (s) where the force is V (s) = U (s) + D(s) (s): Draw the root locus of the angle loop. (c) Compute the transfer function of the new plant from U to Y with D(s) in place. (d) Design a controller Dc (s) for the cart position with the pendulum loop closed. Draw the root locus with respect to the gain of Dc (s) (e) Use MATLAB to plot the control, cart position, and pendulum position for a unit step change in cart position. Solution: (a) U

1 2 s −1

Θ

s 2 − 0.875 s2

Y

5054

CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD s+1 The root locus is shown below. s+9 41 s2 0:875 (c) G1 = 2 s + 8s + 32 s2 2 s + 0:2s + 0:01 (d) Dc = kc : The root locus is shown below. s2 + 2s + 1 (e) The step responses are shown below. The pendulum position control is rather fast for this problem. A more reasonable alternative choice would be to place the pendulum roots at s = 0:5 j0:5: (b) Dp (s) = 41

Inner pendulum loop

Outer cart loop

Final design

2

2

0

-2

1

Amplitude

4

Imag Axis

Imag Axis

1.5 4

0

-2

-4

0.5

0

-4 -0.5

-10

-5

0

-8

-6

-4

-2

0

2

Real Axis

Real Axis

Alternative design

Alternative design

4

0

50

100

150

Time (sec)

1.5 1.2

1

1

Amplitude

Imag Axis

0.5 0 -0.5

0.8 0.6 0.4 0.2

-1 -1.5 -2

0 -0.2 -1

0

1

Real Axis

2

0

20

40

60

80

Time (sec)

Root loci and step responses for Problem 37 38. Consider the 270-ft U.S. Coast Guard cutter Tampa (902) shown in Fig. 5.67. Parameter identi…cation based on sea-trials data (Trankle, 1987) was used to estimate the hydrodynamic coe¢ cients in the equations of motion. The result is that the response of the heading angle of the ship to rudder angle and wind changes w can be described by the second-order transfer functions 0:0184(s + 0:0068) (s) = ; (s) s(s + 0:2647)(s + 0:0063) (s) 0:0000064 Gw (s) = = ; w(s) s(s + 0:2647)(s + 0:0063) G (s) =

5055 where = heading angle, rad r = reference heading angle; rad: r = _ yaw rate; rad=sec; = rudder angle; rad; w = wind speed; m=sec:

Figure 5.67: USCG cutter Tampa (902)

(a) Determine the open-loop settling time of r for a step change in . (b) In order to regulate the heading angle , design a compensator that uses and the measurement provided by a yaw-rate gyroscope (that is, by _ = r). The settling time of to a step change in r is speci…ed to be less than 50 sec, and, for a 5 change in heading the maximum allowable rudder angle de‡ection is speci…ed to be less than 10 . (c) Check the response of the closed-loop system you designed in part (b) to a wind gust disturbance of 10 m=sec (model the disturbance as a step input). If the steady-state value of the heading due to this wind gust is more than 0:5 , modify your design so that it meets this speci…cation as well. Solution:

5056

CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD (a) From the transfer function …nal value theorem, the …nal value is 0.075. Using the step function in MATLAB, the settling time to 1% of the …nal value is ts = 316:11 sec. (b) The maximum de‡ection of the rudder is almost surely at the initial instant, when it is (0) = K r (0): Thus to keep below 10 for a step of 5 ; we need K < 2:and for a settling time less than 50 sec : we need > 4:6=50 = 0:092: Drawing the root locus versus K and using the function rloc…nd we …nd that K = 1:56 gives roots with real parts less than 0:13: The step response shows that this proportional control is adequate for the problem. (c) The steady-state error to a disturbance of 10m= sec is less than 0.35. Open loop step response of variable r 0

Root locus for open loop system 0.92

0.1

-0.02

0.84

0.74

0.6 0.420.22

0.965

Imag Axis

0.05

-0.04

-0.06

0.99 0.25

0

0.2

0.15

0.1

0.05

0.99

-0.05

0.965 -0.1

-0.08

0

100

200

300

0.92

0.84

-0.25

400

-0.2

0.74

-0.15

0.6 0.420.22

-0.1

-0.05

0

Real Axis

Response to heading angle of 5 degrees 0

Response of rudder to 5 degrees 2

-1

1.5

-2 Amplitude

1

-3 -4

0

-5 -6

0.5

-0.5 -1

0

10

20

30

40

50

0

10

20

30

40

Time (sec)

39. Golden Nugget Airlines has opened a free bar in the tail of their airplanes in an attempt to lure customers. In order to automatically adjust for the sudden weight shift due to passengers rushing to the bar when it …rst opens, the airline is mechanizing a pitch-attitude auto pilot. Figure 5.68 shows the block diagram of the proposed arrangement. We will model the passenger moment as a step disturbance Mp (s) = M0 =s, with a maximum expected value for M0 of 0.6. (a) What value of K is required to keep the steady-state error in less than 0.02 rad(= 1 )? (Assume the system is stable.) (b) Draw a root locus with respect to K.

to

5057

Figure 5.68: Golden Nugget Airlines Autopilot

(c) Based on your root locus, what is the value of K when the system becomes unstable? (d) Suppose the value of K required for acceptable steady-state behavior is 600. Show that this value yields an unstable system with roots at s=

2:9; 13:5; +1:2

6:6j:

(e) You are given a black box with rate gyro written on the side and told that when installed, it provides a perfect measure of _ , with output KT _ . Assume K = 600 as in part (d) and draw a block diagram indicating how you would incorporate the rate gyro into the auto pilot. (Include transfer functions in boxes.) (f) For the rate gyro in part (e), sketch a root locus with respect to KT . (g) What is the maximum damping factor of the complex roots obtainable with the con…guration in part (e)? (h) What is the value of KT for part (g)? (i) Suppose you are not satis…ed with the steady-state errors and damping ratio of the system with a rate gyro in parts (e) through (h). Discuss the advantages and disadvantages of adding an integral term and extra lead networks in the control law. Support your comments using MATLAB or with rough root-locus sketches. Solution: (a) K = 300: (b) K = 144

5058

CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD Root locus for problem 5.41

4

3

2

Imag Axis

1

0

-1

-2

-3

-4 -10

-8

-6

-4

-2

0

2

Real Axis

(c) The characteristic equation is s4 + 14s3 + 45s2 + 650s + 1800: The exact roots are 13:5; 2:94; 1:22 6:63: (d) The output of the rate gyro box would be added at the same spot as the attitude sensor output. (e) = 0:28 (f) KT = 185=600 = 0:31 Root Locus 15 0.84

0.74

0.6

0.42

0.22

0.92

10 0.965

5

Imag A xis

0.99

30 0

25

20

15

10

5

0.99 -5

0.965 -10

0.92 0.84 -15 -30

-25

0.74 -20

-15

0.6 -10

0.42

0.22 -5

0

5

10

Real A xis

Root locus for problem 39f (g) Integral (PI) control would reduce the steady-state error to the moment to zero but would make the damping less and the settling time longer. A lead network could improve the damping of the response. 40. Consider the instrument servomechanism with the parameters given in Fig. 5.69. For each of the following cases, draw a root locus with respect to the parameter K, and indicate the location of the roots corresponding to your …nal design. (a) Lead network : Let H(s) = 1;

D(s) = K

s+z ; s+p

p = 6: z

5059

Figure 5.69: Control system for problem 40

Select z and K so that the roots nearest the origin (the dominant roots) yield 0:4;

7;

Kv

2 16 sec 3

1

:

(b) Output-velocity (tachometer) feedback : Let H(s) = 1 + KT s and D(s) = K: Select KT and K so that the dominant roots are in the same location as those of part (a). Compute Kv . If you can, give a physical reason explaining the reduction in Kv when output derivative feedback is used. (c) Lag network : Let H(s) = 1

and D(s) = K

s+1 : s+p

Using proportional control, it is possible to obtain a Kv = 12 at = 0:4. Select K and p so that the dominant roots correspond to the proportional-control case but with Kv = 100 rather than Kv = 12. Solution: (a) The Kv requirement leads to K 55000: With this value, a root locus can be drawn with the parameter z by setting p = 6z:

1+z

6s(s2 + 51s + 550) + K =0 s2 (s2 + 51s + 550) + Ks

5060

CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD

Root locus for Problem 5.40(a)

At the point of maximum damping, the values are z = 17 and the dominant roots are at about 13 j17:

(b) To …nd the values of K and Kv ; we compute a polynomial with roots at 13 j17 and a third pole such that the coe¢ cient of s2 is 51;which is at s = 25:15 This calculation leads to K = 11785, KT = 0:0483 and Kv = 20:81:

(c) The Kv needs to be increased by a factor of 100/12. Thus, we have p = 0:12: The step responses of these designs are given in the plots below.

5061 Stepresponsesforproblem5.42

lag 1.2

lead

A mplitude

1

tach

0.8

0.6

0.4

0.2

0

0

0.2

0.4

0.6

0.8

1

1.2

Time(sec)

Problems and solutions for Section 5.6 41. Plot the loci for the 0 locus or negative K for (a) The examples given in Problem 3 (b) The examples given in Problem 4 (c) The examples given in Problem 5 (d) The examples given in Problem 6 (e) The examples given in Problem 7 (f) The examples given in Problem 8 Solution: 10 plot a 8

6

4

Imag Axis

2

0

-2

-4

-6

-8

(a)

-10 -7

-6

-5

-4

-3

-2 Real Axis

-1

Problem 41(a)

0

1

2

1.4

5062

CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD 10

20

plot a

plot b 10 Imag Axis

Imag Axis

5

0

-5

-10

0

-10

-10

-5 Real Axis

-20

0

1

0

-5 Real Axis

0

plot d 0.5 Imag Axis

0.5 Imag Axis

-5 Real Axis

1 plot c

0

-0.5

-1

-10

0

-0.5

-10

-5 Real Axis

-1

0

-10

Problem 41(b)

5

5 plot b Imag Axis

Imag Axis

plot a

0

-5 -2

-1

0

0

-5 -2

1

-1

Real Axis 5 Imag Axis

Imag Axis

0

1

plot d

0

-5 -2

-1

0

0

-5 -2

1

-1

Real Axis

Real Axis

2

2 plot e

1

Imag Axis

Imag Axis

1

5 plot c

0 -1 -2 -1

0 Real Axis

0

1

2

1

plot f

0 -1 -2 -1

Real Axis

Problem 41(c)

-0.5

0 Real Axis

0.5

1

5063

20

10

0

-20

-10

-5 Real Axis

-10

-5 Real Axis

-8

Imag Axis

20

-4 -2 Real Axis

0

-5 Real Axis

0

-5 Real Axis

0

0

-20

2

0

plot f

Imag Axis

-6

-10

20

0

-5 Real Axis plot d

0

-1

0

plot e

-5

-10

1

plot c

5

Imag Axis

-10

0

0

-20

0

Imag Axis

Imag Axis

20

plot b

Imag Axis

Imag Axis

plot a

-8

-6

-4 -2 Real Axis

0

g 0

-20

-10

Problem 41(d) 20

10

0

-10

0

-5

-10

-5 Real Axis

-10

0

10

-5 Real Axis

-10

0

0.5

Imag Axis

Imag Axis

-10

plot e

5

0

-5

0

-0.5

-10

-5 Real Axis

0

-1 -4

0

-5

1 plot d

-10

plot c

5

Imag Axis

5

Imag Axis

Imag Axis

10

-20

10 plot b

plot a

-2 0 Real Axis

Problem 41(e)

2

-10

-5 0 Real Axis

2

5064

CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD

1

20

1

plot a

plot b

-0.5

0.5

Imag Axis

0

-1

0

-10

-10

-5 Real Axis

-20

0

20

-10

-5 Real Axis

-1 -1

0

10

-10

plot f

0

-5

-10 Real Axis

0

-10 -8

0

-2

-6

-4 -2 0 Real Axis

-4

2

-2

0 Real Axis

Problem 41(f) 42. Suppose you are given the plant L(s) =

3

2

Imag Axis

Imag Axis

0

1 2 Real Axis

4

5

-20

0

plot e

10

Imag Axis

0

-0.5

plot d

-20

plot c

10

Imag Axis

Imag Axis

0.5

1 ; s2 + (1 + )s + (1 + )

where is a system parameter that is subject to variations. Use both positive and negative root-locus methods to determine what variations in can be tolerated before instability occurs. Solution: s+1 : the system is stable for all s2 + s + 1 locus is a circle of radius 1 centered on s = 1: L(s) =

>

1: The complete

43. Consider the system in Fig. 5.70. (a) Use Routh’s criterion to determine the regions in the (K1 ; K2 ) plane for which the system is stable. (b) Use rltool to verify your answer to part (a). Solution: (a) De…ne kp = K1 and kI = K1 K2 and the characteristic equation is s4 + 1:5s3 + 0:5s2 + kp s + kI = 0

2

5065

Figure 5.70: Feedback system for Problem 43

For this equation, Routh’s criterion requires kI > 0; kp < 0:75; and 4kp2 3kp + 9kI < 0: The third of these represents a parabola in the [kp; kI ] plane plotted below. The region of stability is the area under the parabola and above the kp axis. Stability region in the k p,k I plane for problem 5.43 0.07

0.06

0.05

kI

0.04

0.03

0.02

0.01

0

0

0.1

0.2

0.3

0.4 kp

0.5

0.6

0.7

0.8

Stability region for problem 5.43 (b) When kI = 0; there is obviously a pole at the origin. For points on the parabola, consider kp = 3=8 and kI = 1=16: The roots of the characteristic equation are 1:309; 0:191; and j0:5: 44. The block diagram of a positioning servomechanism is shown in Fig. 5.71. (a) Sketch the root locus with respect to K when no tachometer feedback is present (KT = 0). (b) Indicate the root locations corresponding to K = 16 on the locus of part (a). For these locations, estimate the transient-response parameters tr , Mp , and ts . Compare your estimates to measurements obtained using the step command in MATLAB. (c) For K = 16, draw the root locus with respect to KT .

5066

CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD

Figure 5.71: Control system for problem 44

(d) For K = 16 and with KT set so that Mp = 0:05( = 0:707), estimate tr and ts . Compare your estimates to the actual values of tr and ts obtained using MATLAB. (e) For the values of K and KT in part (d), what is the velocity constant Kv of this system? Solution:

(a) The locus is the cross centered at s =

0:5

(b) The roots have a damping of 0.25 and natural frequency of 4. We’d estimate the overshoot to be Mp = 45% and a rise time of less than 0.45 sec. and settling time more than 4.6 sec. The values from the plot are approximately: tr = 0:4; Mp = 45%; and ts = 5 sec : Not too bad. (c) See below. (d) Use rloc…nd on the locus vs Kt to …nd the Kt value that yields 0.7 damping. This shows that KT = 3:66: Using the formulas inside the back cover yields Mp = 0:05; tr = 0:45; and ts = 1:6:

5067

Root locus f or Problem 5.44(a) 4

Step response f or Problem 5.44(b) 1.5

0.860.76 0.64 0.5 0.340.16

2 0.94 Amplitude

Imaginary Axis

3

1 0.985 6 0

5

4

3

2

1

-1 0.985

1

0.5

-2 0.94 -3 -4 -6

0.860.76 0.64 0.5 0.340.16 -4

-2 Real Axis

0 0

2

0

Root locus v s KT, Problem 5.44(c) 4

6

step response f or problem 4.44(d) 1.4

0.860.76 0.64 0.5 0.340.16

3

1.2

2 0.94

1 Amplitude

Imaginary Axis

2 4 Time (sec)

1 0.985 6 0

5

4

3

2

1

0.8 0.6

-1 0.985 0.4

-2 0.94

0.2

-3 -4 -6

0.860.76 0.64 0.5 0.340.16 -4

-2 Real Axis

0 0

2

0

0.5

1 1.5 Time (sec)

Plots for problem 44 (e) Applying Eq. (4.33), we see that Kv = K=(Kt + 2) = 2:83: 45. Consider the mechanical system shown in Fig. 5.72, where g and a0 are gains. The feedback path containing gs controls the amount of rate feedback. For a …xed value of a0 , adjusting g corresponds to varying the location of a zero in the s-plane. (a) With g = 0 and

= 1, …nd a value for a0 such that the poles are

2

2.5

5068

CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD

Figure 5.72: Control system for problem 5.46

complex. (b) Fix a0 at this value, and construct a root locus that demonstrates the e¤ect of varying g. Solution: (a) The roots are complex for a0 > 0:25: We select a0 = 1 and the roots are at s = 0:5 0:866 (b) With respect to g, the root locus equation is s2 + s + 1 + gs = 0: The locus is a circle, plotted below. R ootlocus for problem 5.46

0.8

0.84

0.92

0.74

0.6

0.42

0.22

0.6 0.965 0.4 0.99

Imag Axis

0.2

2 0

1.75

1.5

1.25

1

0.75

0.5

0.25

-0.2 0.99 -0.4 0.965 -0.6

-0.8

-2

0.92

0.84 -1.5

0.74

0.6

-1

0.42 -0.5

0.22 0

R eal A xis

46. Sketch the root locus with respect to K for the system in Fig. 5.73. What is the range of values of K for which the system is unstable? Solution: MATLAB cannot directly plot a root locus for a transcendental function. However, with the Pade’approximation, a locus valid for small values of s can be plotted, as shown below.

5069

Figure 5.73: Control system for problem 5.46

Root locus for problem 5.47 with the (3,3) Pade aproximant 3

2

Imag Axis

1

0

-1

-2

-3 -6

-5

-4

-3

-2

-1

0

Real Axis

Solutions for problem 5.46

1

2

5070

CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD

47. Prove that the plant G(s) = 1=s3 cannot be made unconditionally stable if pole cancellation is forbidden.

Solution:

The angles of departure from a triple pole are 180 and 60 for the negative locus and 0 and 120 for the positive locus. In either case, at least one pole starts out into the right-half plane. Such a system must be conditionally stable for it will be unstable if the gain is small enough.

5071 48. For the equation 1 + KG(s) where G(s) =

1 ; s(s + p)[(s + 1)2 + 4]

use MATLAB to examine the root locus as a function of K for p in the range from p = 1 to p = 10, making sure to include the point p = 2. Solution: The root loci for four values are given in the …gure. The point is that the locus for p = 2 has multiple roots at a complex value of s: Problem 5.49

Problem 5.49 6 4

4 2

Imag Axis

Imag Axis

p= 2

p= 1

6

0 -2 -4

2 0 -2 -4

-6 -6 -5

0

5

-6

-4

0

Real Axis

Problem 5.49

Problem 5.49

10

p= 5

2

4

p = 10

20

5

10

Imag Axis

Imag Axis

-2

Real Axis

0

0

-10

-5

-20 -10 -10

-5

0

5

-20

Real Axis

-10

0 Real Axis

Solutions for problem 48

10

20

Chapter 6

The Frequency-response Design Method Problems and Solutions for Section 6.1 1. (a) Show that

0

in Eq. (6.2) is given by 0

= G(s)

s

U0 ! j!

and 0

= G(s)

=

U0 G( j!)

s= j!

U0 ! s + j!

= U0 G(j!) s=+j!

1 2j

1 : 2j

(b) By assuming the output can be written as y(t) =

0e

j!t

j!t ; 0e

+

derive Eqs. (6.4) - (6.6). Solution: (a) Eq. (6.2):

Y (s) =

1

s

p1

+

2

s

p2

+

+

n

s

pn

+

o

s + j! o

+

o

s

j! o

Multiplying this by (s + j!) :

Y (s)(s+j!) =

1

s + a1

(s+j!)+:::+ 6001

n

s + an

(s+j!)+

o+

o

s

j!

(s+j!)

6002

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

) =

o

= =

o

1

= Y (s)(s + j!)

o js= j!

s + a1

(s + j!) 1

= Y (s)(s + j!)

s + a1

s + an

(s + j!)

j!) o

=

1

(s + j!)

:::

Uo ! (s + j!)js= Y (s)(s + j!)js= j! = G(s) 2 s + !2 Uo ! 1 G(s) js= j! = Uo G( j!) s j! 2j

Similarly, multiplying Eq. (6.2) by (s

Y (s)(s

n

:::

o

s

j!

(s

j!) + ::: +

j!t

j!

n

s + an

(s

o

j!) +

+

j!t oe

y(t)

=

oe

y(t)

=

Uo G( j!)

1 e 2j

G(j!)ej!t

j!t

+ Uo G(j!)

1 j!t e 2j

j!t

G( j!)e 2j

Uo

jG(j!)j

=

\G(j!)

=

n o1 2 2 2 =A Re [G(j!)] + Im [G(j!)]

jG( j!)j

= = =

tan

1

Im [G(j!)] = Re [G(j!)]

n o1 2 2 2 = jG(j!)j Re [G( j!)] + Im [G( j!)] n o 21 2 2 Re [G(j!)] + Im [G(j!)] =A tan

1

Im [G( j!)] = tan Re [G( j!)]

1

) G(j!) = Aej ; G( j!) = Ae

Im [G(j!)] = Re [G(j!)] j

Thus,

y(t)

Aej ej!t

Ae 2j = Uo A sin(!t + ) = Uo

s= j!

(s j!) + s + j! Uo ! = j!)js=j! = G(s) 2 (s j!)js=j! o js=j! = Y (s)(s s + !2 1 Uo ! js=j! = Uo G(j!) = G(s) s + j! 2j s + a1

j

e

(s + j!)

j!) :

=

y(t)

j!

(s + j!)

(b)

\G( j!)

o

s

j!t

= Uo A

ej(!t+

)

e 2j

j(!t+ )

o

6003 where A = jG(j!)j ;

= tan

1

Im [G(j!)] = \G(j!) Re [G(j!)]

2. (a) Calculate the magnitude and phase of

G(s) =

1 s + 10

by hand for ! = 1, 2, 5, 10, 20, 50, and 100 rad/sec. (b) sketch the asymptotes for G(s) according to the Bode plot rules, and compare these with your computed results from part (a). Solution:

(a)

G(s)

1 10 j! 1 ; G(j!) = = s + 10 10 + j! 100 + ! 2 1 ! p ; \G(j!) = tan 1 2 10 100 + !

=

jG(j!)j =

! 1 2 5 10 20 50 100

(b) The Bode plot is :

jG(j!)j 0:0995 0:0981 0:0894 0:0707 0:0447 0:0196 0:00995

\G(j!) 5:71 11:3 26:6 45:0 63:4 78:7 84:3

6004

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Magnitude

10

10

10

10

Bode plot for Problem 6.2

0

-1

-2

-3 -1

10

10

0

1

10 ω (rad/s ec)

10

2

3

10

20

Phase (deg)

0 -20 -40 -60 -80 -100 -1 10

10

0

1

10 ω (rad/s ec)

10

2

3. Sketch the asymptotes of the Bode plot magnitude and phase for each of the following open-loop transfer functions. After completing the hand sketches verify your result using MATLAB. Turn in your hand sketches and the MATLAB results on the same scales. 2000 s(s + 200) 100 (b) L(s) = s(0:1s + 1)(0:5s + 1) 1 (c) L(s) = s(s + 1)(0:02s + 1) 1 (d) L(s) = (s + 1)2 (s2 + 2s + 4) (a) L(s) =

(e) L(s) =

10(s + 4) s(s + 1)(s2 + 2s + 5)

(f) L(s) =

1000(s + 0:1) s(s + 1)(s2 + 8s + 64)

3

10

6005 (g) L(s) =

(s + 5)(s + 3) s(s + 1)(s2 + s + 4)

(h) L(s) =

4s(s + 10) (s + 100)(4s2 + 5s + 4)

(i) L(s) =

s (s + 1)(s + 10)(s2 + 2s + 2500)

Solution:

(a) L(s) =

s

10 +1

s 200

Magnitude

Bode plot for Prob. 6.3 (a)

10

10

0

-5 1

10

10

2

10

3

4

10

ω (rad/s ec) -80

Phase (deg)

-100 -120 -140 -160 -180 -200 1 10

10

2

10 ω (rad/s ec)

(b) L(s) =

s

s 10

100 +1

s 2

+1

3

4

10

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Magnitude

Bode plot for Prob. 6.3 (b)

10

0

-2

10

10

-1

0

10 ω (rad/s ec)

10

1

2

10

-50 -100 Phase (deg)

6006

-150 -200 -250 -300 -2 10

(c) L(s) =

1 s(s+1)(0:02s+1)

10

-1

0

10 ω (rad/s ec)

10

1

2

10

6007

Magnitude

Bode plot for Prob. 6.3 (c)

10

10

0

-5

-2

-1

10

10

0

10

10 ω (rad/s ec)

1

2

10

3

10

-50

Phase (deg)

-100 -150 -200 -250 -300 -2 10

(d) L(s) = (s + 1)2

-1

10

h

1 4 s 2 2

+

s 2

i +1

0

10

10 ω (rad/s ec)

1

2

10

3

10

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Magnitude

10

10

10

Bode plot for Prob. 6.3 (d)

0

-2

-4 -1

0

10

10 ω (rad/s ec)

1

10

0 Phase (deg)

6008

-100 -200 -300 -1

0

10

10 ω (rad/s ec)

8

(e) L(s) = s(s + 1)

s 4

+1

ps 5

2

+ 25 s + 1

1

10

6009

Magnitude

10

10

10

Bode plot for Prob. 6.3 (e)

5

0

-5 -2

10

10

-1

0

10 ω (rad/s ec)

10

1

2

10

-50

Phase (deg)

-100 -150 -200 -250 -300 -350 -2

10

(f) L(s) = s(s +

10

25 16 h (10s + 1) 2 1) 8s + 8s +

-1

i 1

0

10 ω (rad/s ec)

10

1

2

10

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Magnitude

10

10

Bode plot for Prob. 6.3 (f)

5

0

-3

-2

10

10

10

-1

10 ω (rad/s ec)

0

1

10

2

10

0 Phase (deg)

6010

-100 -200 -300 -3

-2

10

(g) L(s) =

15 4

s(s +

10

s 5h+ 1 2 1) 2s

s 3

+

+1 s 4

i +1

10

-1

10 ω (rad/s ec)

0

1

10

2

10

6011

Magnitude

Bode plot for Prob. 6.3 (g)

10

0

-2

10

10

-1

0

10 ω (rad/s ec)

10

1

2

10

0

Phase (deg)

-100 -200 -300 -400 -2

10

(h) L(s) =

10

1 10 s 100

+1

s

s 10 + 1 s2 + 54 s

-1

+1

0

10 ω (rad/s ec)

10

1

2

10

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Bode plot for Prob. 6.3 (h)

Magnitude

10

10

10

-1

-2

-3 -2

-1

10

0

10

10

10 ω (rad/s ec)

1

2

10

3

10

100 Phase (deg)

6012

0 -100 -2

-1

10

(i) L(s) = (s + 1)

0

10

s 10

+

1 s 25000 h s 2 1 50

10

+

1 1250 s

10 ω (rad/s ec)

i +1

1

2

10

3

10

6013

Magnitude

Bode plot for Prob. 6.3 (i)

10

-5

-2

-1

10

0

10

10

10 ω (rad/s ec)

1

2

10

3

10

Phase (deg)

100 0 -100 -200 -300 -2

-1

10

0

10

10

10 ω (rad/s ec)

1

4. Real poles and zeros. Sketch the asymptotes of the Bode plot magnitude and phase for each of the following open-loop transfer functions. After completing the hand sketches verify your result using MATLAB. Turn in your hand sketches and the MATLAB results on the same scales. (a) L(s) =

1 s(s + 1)(s + 5)(s + 10)

(b) L(s) =

(s + 2) s(s + 1)(s + 5)(s + 10)

(c) L(s) =

(s + 2)(s + 6) s(s + 1)(s + 5)(s + 10)

(d) L(s) =

(s + 2)(s + 4) s(s + 1)(s + 5)(s + 10)

Solution: (a) L(s) =

1 50

s(s + 1)

s 5

+1

s 10

+1

2

10

3

10

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Bode plot for Prob. 6.4 (a) 0

Magnitude

10

-10

10

-2

-1

10

0

10

10

10 ω (rad/s ec)

1

2

10

3

10

0 -100 Phase (deg)

6014

-200 -300 -400 -2

-1

10

(b) L(s) =

0

10

s 2

1 25

s(s + 1)

s 5

+1 +1

s 10

1

10 10 Frequency (rad/s ec)

+1

2

10

3

10

6015 Bode plot for Prob. 6.4 (b) 0

Magnitude

10

-10

10

-2

-1

10

10

0

10

10 ω (rad/s ec)

1

2

10

3

10

Phase (deg)

0 -100 -200 -300 -2

-1

10

(c) L(s) =

10

+ 1 6s + 1 s s(s + 1) 5s + 1 10 +1 6 25

s 2

0

10

10 ω (rad/s ec)

1

2

10

3

10

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Bode plot for Prob. 6.4 (c) 0

Magnitude

10

-5

10

-10

10

-2

-1

10

10

0

10

10 ω (rad/s ec)

1

2

10

3

10

0 -50 Phase (deg)

6016

-100 -150 -200 -250 -2

-1

10

(d) L(s) =

10

+ 1 4s + 1 s s(s + 1) 5s + 1 10 +1 4 25

s 2

0

10

10 ω (rad/s ec)

1

2

10

3

10

6017 Bode plot for Prob. 6.4 (d) 0

Magnitude

10

-5

10

-10

10

-2

10

-1

10

0

10

10 ω (rad/s ec)

1

2

10

3

10

Phase (deg)

0 -50 -100 -150 -200 -250 -2

10

-1

10

0

10

10 ω (rad/s ec)

1

5. Complex poles and zeros Sketch the asymptotes of the Bode plot magnitude and phase for each of the following open-loop transfer functions and approximate the transition at the second order break point based on the value of the damping ratio. After completing the hand sketches verify your result using MATLAB. Turn in your hand sketches and the MATLAB results on the same scales. 1 + 3s + 10 1 (b) L(s) = s(s2 + 3s + 10) (a) L(s) =

s2

(c) L(s) =

(s2 + 2s + 8) s(s2 + 2s + 10)

(d) L(s) =

(s2 + 2s + 12) s(s2 + 2s + 10)

(e) L(s) =

(s2 + 1) s(s2 + 4)

2

10

3

10

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD (s2 + 4) s(s2 + 1)

(f) L(s) =

Solution:

1 10

(a) L(s) = ps 10

2

+

3 10 s

+1

Bode plot for Prob. 6.5 (a)

0

Magnitude

10

10

-5 -1

10

10

0

10

1

2

10

ω (rad/s ec) 50 0 Phase (deg)

6018

-50 -100 -150 -200 -1

10

10

0

10 ω (rad/s ec)

(b) L(s) = s

ps 10

1 10 2

+

3 10 s

+1

1

2

10

6019 Bode plot for Prob. 6.5 (b) 0

Magnitude

10

-5

10

-10

10

-1

10

10

0

10

1

2

10

ω (rad/s ec)

Phase (deg)

-50 -100 -150 -200 -250 -300 -1

10

10

0

10 ω (rad/s ec)

4 5

s p

2 2

(c) L(s) = s

2

ps 10

2

+ 14 s + 1 + 15 s + 1

1

2

10

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Magnitude

Bode plot for Prob. 6.5 (c)

10

0

-1

10

10

0

10

1

2

10

ω (rad/s ec)

0 Phase (deg)

6020

-50 -100 -150 -1 10

10

0

10 ω (rad/s ec)

6 5

s p

2 3

(d) L(s) = s

2

ps 10

2

+ 16 s + 1 + 15 s + 1

1

2

10

6021

Magnitude

Bode plot for Prob. 6.5 (d)

10

10

0

-2

-1

10

10

0

10

1

2

10

ω (rad/s ec)

Phase (deg)

-50 -100 -150 -200 -1

10

10

0

10 ω (rad/s ec)

1

(e) L(s) = s

h4

(s2 + 1) s 2 2

i +1

1

2

10

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Magnitude

Bode plot for Prob. 6.5 (e)

10

10

0

-2

-1

0

10

10 ω (rad/s ec)

1

10

150 100 Phase (deg)

6022

50 0 -50 -100 -150 -1 10

4 (f) L(s) =

h

s 2 2

0

10 ω (rad/s ec)

i +1

s(s2 + 1)

1

10

6023

Magnitude

Bode plot for Prob. 6.5 (f)

10

0

-1

10

0

10 ω (rad/s ec)

1

10

150

Phase (deg)

100 50 0 -50 -100 -150 -1 10

0

10 ω (rad/s ec)

6. Multiple poles at the origin Sketch the asymptotes of the Bode plot magnitude and phase for each of the following open-loop transfer functions. After completing the hand sketches verify your result using MATLAB. Turn in your hand sketches and the MATLAB results on the same scales.

(a) L(s) = (b) L(s) = (c) L(s) = (d) L(s) = (e) L(s) = (f) L(s) =

1 + 8) 1 s3 (s + 8) 1 s4 (s + 8) (s + 3) s2 (s + 8) (s + 3) s3 (s + 4) (s + 1)2 s3 (s + 4) s2 (s

1

10

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD (g) L(s) =

(s + 1)2 + 10)2

s3 (s

Solution:

(a) L(s) =

1 8 s 8 +

s2

1

Bode plot for Prob. 6.6 (a)

Magnitude

10

10

0

-5

-1

10

10

0

10

1

2

10

ω (rad/s ec)

-150 Phase (deg)

6024

-200 -250 -300 -1

10

10

0

10 ω (rad/s ec)

(b) L(s) =

s3

1 8 s 8 +

1

1

2

10

6025 Bode plot for Prob. 6.6 (b) 0

Magnitude

10

-10

10

-1

10

10

0

10

1

2

10

ω (rad/s ec)

Phase (deg)

-250 -300 -350 -400 -1

10

10

0

10 ω (rad/s ec)

(c) L(s) =

s4

1 8 s 8 +

1

1

2

10

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Bode plot for Prob. 6.6 (c)

Magnitude

0

10

-10

10

-1

10

10

0

10

1

2

10

ω (rad/s ec) 50 Phase (deg)

6026

0 -50 -100 -150 -1 10

10

0

10 ω (rad/s ec)

(d) L(s) =

3 s 8 3 +1 s2 8s + 1

1

2

10

6027

Magnitude

10

10

10

Bode plot for Prob. 6.6 (d)

5

0

-5 -1

10

10

0

10

1

2

10

ω (rad/s ec)

Phase (deg)

-50 -100 -150 -200 -1

10

10

0

10 ω (rad/s ec)

(e) L(s) =

3 s 4 3 +1 s3 4s + 1

1

2

10

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Magnitude

10

10

10

Bode plot for Prob. 6.6 (d)

5

0

-5 -1

10

10

0

10

1

2

10

ω (rad/s ec) -50 Phase (deg)

6028

-100 -150 -200 -1

10

10

0

10 ω (rad/s ec)

(f) L(s) =

2 1 4 (s + 1) s3 4s + 1

1

2

10

6029 Bode plot for Prob. 6.6 (f)

Magnitude

10

10

10

5

0

-5 -2

10

10

-1

0

10 ω (rad/s ec)

10

1

2

10

Phase (deg)

-50 -100 -150 -200 -250 -300 -2

10

(g) L(s) =

1 100 (s s s3 10

10

2

+ 1) +1

2

-1

0

10 ω (rad/s ec)

10

1

2

10

6030

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Magnitude

10

10

10

Bode plot for Prob. 6.6 (g)

5

0

-5

-2

10

10

-1

0

10 ω (rad/s ec)

10

1

2

10

Phase (deg)

-50 -100 -150 -200 -250 -300 -2

10

10

-1

0

10 ω (rad/s ec)

10

1

7. Mixed real and complex poles Sketch the asymptotes of the Bode plot magnitude and phase for each of the following open-loop transfer functions. After completing the hand sketches verify your result using Matlab. Turn in your hand sketches and the Matlab results on the same scales. (a) L(s) =

(s + 2) s(s + 10)(s2 + 2s + 2)

(b) L(s) =

(s + 2) s2 (s + 10)(s2 + 6s + 25)

(c) L(s) =

(s + 2)2 s2 (s + 10)(s2 + 6s + 25)

(d) L(s) =

(s + 2)(s2 + 4s + 68) s2 (s + 10)(s2 + 4s + 85)

(e) L(s) = Solution:

[(s + 1)2 + 1] + 2)(s + 3)

s2 (s

2

10

6031 1 10

(a) L(s) = s

s 10

+1

s 2

+1

ps 2

2

+s+1

Bode plot for Prob. 6.7 (a)

Magnitude

10

10

0

-5

-2

10

10

-1

0

10 ω (rad/s ec)

10

1

2

10

Phase (deg)

-50 -100 -150 -200 -250 -300 -2

10

(b) L(s) = s2

s 10

10

+

-1

1 s 125h 2 + 1 s 2 6 1 + 25 s 5

0

10 ω (rad/s ec)

i +1

10

1

2

10

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Bode plot for Prob. 6.7 (b)

Magnitude

0

10

-5

10

-10

10

-2

10

10

-1

0

10 ω (rad/s ec)

10

1

2

10

-100 Phase (deg)

6032

-200 -300 -400 -2

10

(c) L(s) = s2

10

-1

2 2 s 125 h 2 + 1 s s 2 6 + 25 s 10 + 1 5

0

10 ω (rad/s ec)

i +1

10

1

2

10

6033

Magnitude

Bode plot for Prob. 6.7 (c)

10

10

0

-5

-2

10

10

-1

0

10 ω (rad/s ec)

10

1

2

10

Phase (deg)

0 -100 -200 -300 -2

10

4 25

s 2

s2

s 10

10

+1

ps 2 17

(d) L(s) = +1

ps 85

-1

2

0

10 ω (rad/s ec)

+

1 17 s

+1

+

4 85 s

+1

2

10

1

2

10

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Magnitude

10

10

10

Bode plot for Prob. 6.7 (d)

5

0

-5 -2

10

10

-1

0

10 ω (rad/s ec)

10

1

2

10

100 Phase (deg)

6034

0 -100 -200 -2

10

(e) L(s) =

10

1 3

ps 2

s2

s 2

2

+1

+s+1 s 3

+1

-1

0

10 ω (rad/s ec)

10

1

2

10

6035

Magnitude

10

10

10

Bode plot for Prob. 6.7 (e)

5

0

-5 -2

10

10

-1

0

10 ω (rad/s ec)

10

1

2

10

Phase (deg)

0 -50 -100 -150 -200 -2

10

10

-1

0

10 ω (rad/s ec)

10

1

8. Right half plane poles and zeros Sketch the asymptotes of the Bode plot magnitude and phase for each of the following open-loop transfer functions. Make sure the phase asymptotes properly take the RHP singularity into account by sketching the complex plane to see how the \L(s) changes as s goes from 0 to +j1: After completing the hand sketches verify your result using MATLAB. Turn in your hand sketches and the MATLAB results on the same scales. s+2 1 ; The model for a case of magnetic levitation s + 10 s2 1 with lead compensation. s+2 1 (b) L(s) = ; The magnetic levitation system with ins(s + 10) (s2 1) tegral control and lead compensation. s 1 (c) L(s) = s2 s2 + 2s + 1 (d) L(s) = s(s + 20)2 (s2 2s + 2) (a) L(s) =

2

10

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD (e) L(s) =

(f) L(s) =

(s + 2) s(s 1)(s + 6)2 1 1)[(s + 2)2 + 3]

(s

Solution:

(a) L(s) =

1 5

s 2

+1 1 s + 10 s2 1

Bode plot for Prob. 6.8 (a) 0

Magnitude

10

10

-5 -1

10

10

0

10

1

2

10

ω (rad/s ec)

-100 Phase (deg)

6036

-150 -200 -1

10

10

0

10 ω (rad/s ec)

(b) L(s) =

1 5

s 2

+1 1 s(s + 10) s2 1

1

2

10

6037 Bode plot for Prob. 6.8 (b)

Magnitude

10

10

0

-5

-2

10

10

-1

0

10 ω (rad/s ec)

10

1

2

10

Phase (deg)

-150 -200 -250 -300 -2

10

(c) L(s) =

s

1 s2

10

-1

0

10 ω (rad/s ec)

10

1

2

10

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Magnitude

10

10

Bode plot for Prob. 6.8 (c)

5

0

-2

10

10

-1

10

0

1

10

ω (rad/s ec) 50 Phase (deg)

6038

0 -50 -100 -150 -2 10

10

-1

10 ω (rad/s ec)

1 2 40 (s

(d) L(s) = s

s 20

+1

2

+ 2s + 1) ps 2

2

s+1

0

1

10

6039

Magnitude

10

10

Bode plot for Prob. 6.8 (d)

0

-5

-2

10

10

-1

0

10 ω (rad/s ec)

10

1

2

10

300

Phase (deg)

200 100 0 -100 -2

10

1 18

(e) L(s) = s(s

10

s 2

1)

+1 s 6

+1

2

-1

0

10 ω (rad/s ec)

10

1

2

10

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Bode plot for Prob. 6.8 (e)

Magnitude

10

10

0

-5

-2

10

10

-1

0

10 ω (rad/s ec)

10

1

2

10

-100 Phase (deg)

6040

-150 -200 -250 -300 -2

10

10

-1

1 7

(f) L(s) = (s

1)

ps 7

2

+ 47 s + 1

0

10 ω (rad/s ec)

10

1

2

10

6041

Figure 6.87: Magnitude portion of Bode plot for Problem 9

Bode plot for Prob. 6.8 (f)

Magnitude

10

10

0

-5

-2

10

10

-1

0

10 ω (rad/s ec)

10

1

2

10

Phase (deg)

-100 -150 -200 -250 -300 -2

10

10

-1

0

10 ω (rad/s ec)

10

1

9. A certain system is represented by the asymptotic Bode diagram shown in Fig. 6.88. Find and sketch the response of this system to a unit step input (assuming zero initial conditions). Solution: By inspection, the given asymptotic Bode plot is from

2

10

6042

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Therefore, G(s) =

10(s=10 + 1) s + 10 = s s

The response to a unit step input is : Y (s)

= G(s)U (s) s + 10 1 1 10 = = + 2 s s s s 1 y(t) = $ [Y (s)] = 1(t) + 10t (t 0)

Prob. 6.9: Unit Step Res pons e 25

20

y(t)

15

10

5

0

0

0.5

1 Tim e (sec)

10. Prove that a magnitude slope of per decade or -6 db per octave.

1.5

1 in a Bode plot corresponds to

Solution: The de…nition of db is db = 20 log jGj Assume slope =

d(logjGj) d(log !)

=

1

(1) (2)

2

20 db

6043 (2) =) log jGj =

log ! + c (c is a constant.)

(1) and (3) =) db =

(3)

20 log ! + 20c

Di¤erentiating this, d (db) = d (log !)

20

Thus, a magnitude slope of -1 corresponds to -20 db per decade. Similarly, d (db) d (db) = ! d (log2 !) d log log 2

+

6

Thus, a magnitude slope of -1 corresponds to -6 db per octave. 11. A normalized second-order system with a damping ratio additional zero is given by G(s) =

= 0:5 and an

s=a + 1 : s2 + s + 1

Use MATLAB to compare the Mp from the step response of the system for a = 0:01; 0:1, 1, 10, and 100 with the Mr from the frequency response of each case. Is there a correlation between Mr and Mp ? Solution:

0:01 0:1 1 10 100

Resonant peak, Mr 98:8 9:93 1:46 1:16 1:15

Overshoot, Mp 54:1 4:94 0:30 0:16 0:16

As is reduced, the resonant peak in frequency response increases. This leads us to expect extra peak overshoot in transient response. This e¤ect is signi…cant in case of = 0:01; 0:1; 1, while the resonant peak in frequency response is hardly changed in case of = 10. Thus, we do not have considerable change in peak overshoot in transient response for 10. The response peak in frequency response and the peak overshoot in transient response are correlated.

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Unit Step Res pons e 60

50 a=0.01 40

30 y(t)

6044

20

10

a=0.1 a=1

a=100

0

-10

a=10

0

2

4

6 Tim e (s ec)

8

10

6045 Bode plot for Prob. 6.11

Magnitude

a=0.01 10

a=0.1 a=1

0

a=10 a=100 10

-5 -2

10

Phase (deg)

100

10

-1

0

10 ω (rad/s ec)

10

1

2

10

a=0.01 a=0.1 a=1

0

-100

a=10 a=100

-200 -2 10

10

-1

12. A normalized second-order system with given by. G(s) =

0

10 ω (rad/s ec)

10

1

= 0:5 and an additional pole is

1 [(s=p) + 1](s2 + s + 1)

Draw Bode plots with p = 0:01; 0:1, 1, 10 and 100. What conclusions can you draw about the e¤ect of an extra pole on the bandwidth compared to the bandwidth for the second-order system with no extra pole? Solution: p 0:01 0:1 1 10 100

Additional pole ( p) 0:01 0:1 1 10 100

Bandwidth, ! Bw 0:013 0:11 1:0 1:5 1:7

As p is reduced, the bandwidth decreases. This leads us to expect slower time response and additional rise time. This e¤ect is signi…cant in

2

10

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD case of p = 0:01; 0:1; 1, while the bandwidth is hardly changed in case of p = 10. Thus, we do not have considerable change in rise time for p 10.

Bandwidth is a measure of the speed of response of a system, such as rise time.

Unit Step Res pons e 1.4 a=100

1.2

1 a=10 0.8 y(t)

6046

a=0.1

a=1 0.6

0.4

0.2

0

a=0.01

0

5

10

15

20 Tim e (s ec)

25

30

35

40

6047 Bode plot for Prob. 6.12 10

0

Magnitude

a=0.1 a=1

a=100 a=10

a=0.01 10

-5

-2

10

10

-1

0

10 Frequency (rad/s ec)

10

1

2

10

0 Phase (deg)

a=0.1 -100

a=1

a=0.01 a=100

-200 -300 -2 10

a=10

10

-1

0

10 ω (rad/s ec)

10

1

13. For the closed-loop transfer function T (s) =

! 2n ; s2 + 2 ! n s + ! 2n

derive the following expression for the bandwidth ! BW of T (s) in terms of ! n and : ! BW = ! n

r

1

2

Assuming ! n = 1, plot ! BW for 0

2

+

q

2+4

1.

Solution : The closed-loop transfer function : T (s) =

s2

! 2n + 2 ! n s + ! 2n

4

4 2:

2

10

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD s = j!; T (j!)

1

=

2

! !n

1

! !n

+2

j

2

1 6 jT (j!)j = fT (j!)T (j!)g 2 = 4

Let x =

= =) =)

"

1 2

! !n

1

1 (1

2

x2 ) + (2 x)2

x4 + (4

2

2)x2

! BW x= = (1 !n r

=) ! BW = ! n

2

0:2 0:5 0:8

1

# 21

7 o2 5

! !n

1 = 0:707 = p 2

q 2 ) + (1 2

2

! BW !n 1:51 1:27 0:87

2

q + 2+4 ! BW

1:51! n 1:27! n 0:87! n

Bandwidth vs zeta 1.6

BW

1.4 1.2 1 0.8

0

n + 2

1=0

x =

0.6

3 12

! BW : !n

jT (j!)j!=!BW

Bandwidth, ω

6048

0.2

0.4 0.6 Dam ping ratio,ζ

0.8

1

2 4

2 2 )

4

2

1 2

+1

6049 14. Consider the system whose transfer function is

G(s) =

A0 ! 0 s : Qs2 + ! 0 s + ! 20 Q

This is a model of a tuned circuit with quality factor Q. (a) Compute the magnitude and phase of the transfer function analytically, and plot them for Q = 0:5, 1, 2, and 5 as a function of the normalized frequency !=! 0 . (b) De…ne the bandwidth as the distance between the frequencies on either side of ! 0 where the magnitude drops to 3 db below its value at ! 0 and show that the bandwidth is given by

BW =

!0 Q

1 2

:

(c) What is the relation between Q and ? Solution :

(a) Let s = j!;

G(j!)

= =

jG(j!)j =

=

The normalized magnitude normalized frequency

! !o

Ao ! o j! + ! o j! + ! 2o Q Ao

Q! 2 1+

Q! 2o Q! 2 j! o !

Ao

r

1 + Q2

tan

QG(j!) Ao

1

! !o

! !o

!o !

2

!o !

and phase are plotted against

for di¤erent values of Q.

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

10

Bode plot for Prob. 6.14

1

QG(jω)/A

o

Q=5

10

Q=2

0

Q=1 Q=0.5 10

-1 -1

0

10

10 ω/ω

1

10

o

100

Phase (deg)

6050

50 0

Q=0.5 Q=1 Q=2

-50 Q=5 -100 -1 10

0

10 ω/ω

1

10

o

(b) There is symmetry around ! o . For every frequency ! 1 < ! o , there exists a frequency ! 2 > ! o which has the same magnitude jG(j! 1 )j = jG(j! 2 )j We have that, !1 !o

!o = !1

!2 !o

!o !2

which implies ! 2o = ! 1 ! 2 . Let ! 1 < ! o and ! 2 > ! o be the two frequencies on either side of ! o for which the gain drops by 3db from its value of Ao at ! o . BW =

!2

!1 2

=

1 2

!2

! 2o !2

Now ! 2 is found from, G(j!) 1 =p Ao 2

(1)

6051 or 1 + Q2

!2 !o

!o !2

Q !o

!2

2

=2

which yields Q

!2 !o

!o !2

=1=

! 2o !2

(2)

Comparing (1) and (2) we …nd, 1 BW = p 2

Q !o

(c) G(s)

= =

Qs2

A0 ! 0 s + ! 0 s + ! 20 Q A0 ! 0 s

Q s2 + = Therefore

!0 Qs

+ ! 20

A0 ! 0 s Q (s2 + 2 ! 0 s + ! 20 ) 1 =2 Q

15. A DC voltmeter schematic is shown in Fig. 6.88. The pointer is damped so that its maximum overshoot to a step input is 10%. (a) What is the undamped natural frequency of the system? (b) What is the damped natural frequency of the system? (c) Plot the frequency response using MATLAB to determine what input frequency will produce the largest magnitude output? (d) Suppose this meter is now used to measure a 1-V AC input with a frequency of 2 rad/sec. What amplitude will the meter indicate after initial transients have died out? What is the phase lag of the output with respect to the input? Use a Bode plot analysis to answer these questions. Use the lsim command in MATLAB to verify your answer in part (d). Solution : The equation of motion : I • + b _ + k = T = Km v, where b is a damping coe¢ cient.

6052

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Figure 6.88: Voltmeter schematic

Taking the Laplace transform with zero initial conditions: (s) =

Use I = 40

10

6

Km Km I V (s) = V (s) Is2 + bs + k s2 + 2 ! n s + ! 2n

Kg m2 , k = 4

6

10

Kg m2 /s2 , Km =4 10

6

N m/v

(a) Undamped natural frequency: k = =) ! n = I

! 2n

(b) Since Mp = 0:1 and Mp = e log 0:1 = p 1

2

p

=)

1

r 2

k = 0:316 rad/sec I

;

= 0:5911 (' 0:6 from Figure 2.44)

Damped natural frequency: !d = !n

q 1

2

= 0:255 rad/sec

6053 (c)

T (j!)

=

jT (j!)j = d jT (j!)j d!

When

djT (j!)j d!

=

(j!) Km =I = V (j!) (j!)2 + 2 ! n j! + ! 2n Km =I [(! 2n Km I

1

! 2 )2 + (2 ! n !)2 ] 2 2! ! 2n [(! 2n

!2

2 2 ! 2n 3

! 2 )2 + (2 ! n !)2 ] 2

= 0;

!2

(1

2 2 )! 2n !

= =

0 0:549! n = 0:173

Alternatively, the peak frequency can be found from the Bode plot:

! = 0:173 rad/sec

(d) With ! = 2 rad/sec from the Bode plot:

Amplitude Phase

= 0:0252 rad = 169:1

6054

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Bode plot for Prob. 6.15 0

Magnitude

10

|G(jω)|=0.025 10

ω=2

-2

-1

10

0

1

10 ω (rad/s ec)

10

Phase (deg)

0 -50 -100 -150

-169 (deg) ω=2

-200 -1 10

0

10 ω (rad/s ec)

1

10

6055

Problems and Solutions for Section 6.2 16. Determine the range of K for which the closed-loop systems (see Fig. 6.18) are stable for each of the cases below by making a Bode plot for K = 1 and imagining the magnitude plot sliding up or down until instability results. Verify your answers using a very rough sketch of a root-locus plot. K(s + 2) s + 20 K (b) KG(s) = (s + 10)(s + 1)2 K(s + 10)(s + 1) (c) KG(s) = (s + 100)(s + 5)3 (a) KG(s) =

Solution : (a) KG(s) =

K(s + 2) K = s + 20 10

s 2 +1 s 20 + 1

Bode plot for Prob. 6.16 (a) -0.1

-0.5

10

-0.9

10

-1

10

10

0

1

10 ω (rad/s ec)

10

2

3

10

60

Phase (deg)

Magnitude

10

40

20

0 -1 10

10

0

1

10 ω (rad/s ec)

10

2

3

10

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Root Locus 1

0.8

0.6

0.4

Imaginary Axis

6056

0.2

0

-0.2

-0.4

-0.6

-0.8

-1 -25

-20

-15

-10 Real Axis

-5

The gain can be raised or lowered on the Bode gain plot and the phase will never be less than -180o , so the system is stable for any K > 0:

(b)

KG(s) =

K K = (s + 10)(s + 1)2 10

s 10

1 + 1 (s + 1)2

0

5

6057 Bode plot for Prob. 6.16(b)

0

10

Magnitude

K=1/242 -5

10

-10

10

-1

10

10

0

10

1

2

10

ω (rad/s ec)

Phase (deg)

0

-100

-200

-300 -1 10

10

0

10

1

ω (rad/s ec)

The bode plots show that the gain, K, would equal 242 when the phase crosses 180o : So, K < 242 is Stable and K > 242 is Unstable. The phase crosses the 180o at ! = 4:58 rad/sec. The root locus below veri…es the situation.

2

10

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Root Locus 25

20

15

10

Imaginary Axis

6058

5

0

-5

-10

-15

-20

-25 -30

-25

-20

-15

-10 Real Axis

-5

0

(c)

KG(s) =

K(s + 10)(s + 1) K = 3 (s + 100)(s + 5) 1250

s 10 + 1 s 100 + 1

(s + 1) s 5

+1

3

5

10

6059

Magnitude

10

10

10

Bode plot for Prob. 6.16 (c)

-2

-4

-6 -1

10

10

0

1

10 ω (rad/s ec)

10

2

3

10

50

Phase (deg)

0 -50 -100 -150 -200 -1 10

10

0

1

10 ω (rad/s ec)

10

2

The phase never crosses -180o so it is stable for all K > 0; as con…rmed by the root locus. 17. Determine the range of K for which each of the following systems is stable by making a Bode plot for K = 1 and imagining the magnitude plot sliding up or down until instability results. Verify your answers using a very rough sketch of a root-locus plot. (a) KG(s) =

K(s + 1) s(s + 5)

K(s + 1) s2 (s + 10) K (c) KG(s) = (s + 2)(s2 + 9)

(b) KG(s) =

(d) KG(s) = Solution :

K(s + 1)2 s3 (s + 10)

3

10

6060

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Root Locus 30

20

Imaginary Axis

10

0

-10

-20

-30 -120

-100

-80

-60

-40 Real Axis

-20

0

20

6061 (a)

KG(s) =

Magnitude

10

10

10

10

K(s + 1) K (s + 1) = s(s + 5) 5 s 5s + 1

Bode plot for Prob. 6.17 (a)

1

0

-1

-2 -1

10

10

0

10

1

2

10

ω (rad/s ec) -40

Phase (deg)

-50 -60 -70 -80 -90 -1 10

10

0

10

1

ω (rad/s ec)

The phase never crosses -180o so it is stable for all K > 0; as con…rmed by the root locus.

2

10

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Root Loc us 0.8

0.6

0.4

0.2 Imaginary Axis

6062

0

-0.2

-0.4

-0.6

-0.8 -16

-14

-12

-10

-8

-6 Real Axis

(b)

KG(s) =

K(s + 1) K s+1 = s s2 (s + 10) 10 s2 10 +1

-4

-2

0

2

6063

Magnitude

10

10

10

10

Bode plot for Prob. 6.17(b)

2

0

-2

-4 -1

10

10

0

10

1

2

10

ω (rad/s ec)

Phase (deg)

-120

-140

-160

-180 -1 10

10

0

10

1

ω (rad/s ec)

The phase never crosses -180o so it is stable for all K > 0; as con…rmed by the root locus. The system is stable for any K > 0:

(c)

KG(s) =

K K = 2 (s + 2)(s + 9) 18

1 s 2

+1

s2 9

+1

2

10

6064

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Root Locus 8

6

4

Imaginary Axis

2

0

-2

-4

-6

-8 -12

-10

-8

-6

-4 Real Axis

-2

0

2

6065 Bode plot for Fig. 6.17 (c)

5

Magnitude

10

0

10

-5

10

-10

10

-1

10

10

0

10

1

2

10

ω (rad/s ec) 150

Phase (deg)

100 50 0 -50 -100 -1 10

10

0

10

1

ω (rad/s ec)

The bode is di¢ cult to read, but the phase really dropped by 180o at the resonance. (It appears to rise because of the quadrant action in Matlab) Furthermore, there is an in…nite magnitude peak of the gain at the resonance because there is zero damping. That means that no matter how much the gain is lowered, the gain will never cross magnitude one when the phase is -180o . So it can not be made stable for any K. This is much clearer and easier to see in the root locus below.

2

10

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Root Locus 15

10

5

Imaginary Axis

6066

0

-5

-10

-15 -20

-15

-10

-5 Real Axis

(d)

KG(s) =

K(s + 1)2 K (s + 1)2 = s 3 s (s + 10) 10 s3 10 +1

0

5

10

6067

Magnitude

10

10

10

Bode plot for Prob. 6.17 (d)

5

0

Gain=1/6.17

-5 -1

10

10

0

10

1

2

10

ω (rad/s ec)

Phase (deg)

400 200 0 -200 -400 -1 10

10

0

10

1

ω (rad/s ec)

This is not the normal situation discussed in Section 6.2 where increasing gain leads to instability. Here we see from the root locus that K must be 6.17 in order for stability. Note that the phase is increasing with frequency here rather than the normal decrease we saw on the previous problems. It’s also interesting to note that the margin command in Matlab indicates instability! (which is false.) This problem illustrates that a sketch of the root locus really helps understand what’s going on... and that you can’t always trust Matlab, or at least that you need good understanding to interpret what Matlab is telling you.

2

10

6068

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

K < 6:25 : Unstable

K > 6:25 : Stable

! = 1:12 rad/sec for K = 6:17.

6069

Problems and Solutions for Section 6.3 18. (a) Sketch the Nyquist plot for an open-loop system with transfer function 1=s2 ; that is, sketch 1 js=C1 ; s2 where C1 is a contour enclosing the entire RHP, as shown in Fig. 6.17. (Hint: Assume C1 takes a small detour around the poles at s = 0, as shown in Fig. 6.27.) (b) Repeat part (a) for an open-loop system whose transfer function is G(s) = 1=(s2 + ! 20 ). Solution : (a) 1 s2 Note that the portion of the Nyquist diagram on the right side below that corresponds to the bode plot is from B’to C’. The large loop from F’to A’to B’arises from the detour around the 2 poles at the origin. G(s) =

(b) G(s) =

1 s2 + ! 20

Note here that the portion of the Nyquist plot coming directly from a Bode plot is the portion from A’ to E’. That portion includes a 180o arc that arose because of the detour around the pole on the imaginary axis.

6070

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

19. Sketch the Nyquist plot based on the Bode plots for each of the following systems, then compare your result with that obtained using the MATLAB command nyquist:

(a) KG(s) =

K(s + 2) s + 10

(b) KG(s) =

K (s + 10)(s + 2)2

(c) KG(s) =

K(s + 10)(s + 1) (s + 100)(s + 2)3

(d) Using your plots, estimate the range of K for which each system is stable, and qualitatively verify your result using a rough sketch of a root-locus plot.

Solution :

(a)

6071

N = 0; P = 0 =) Z = N + P = 0 The closed-loop system is stable for any K > 0: (b) The Bode plot shows an initial phase of 0o hence the Nyquist starts on the positive real axis at A’. The Bode ends with a phase of 270o hence the Nyquist ends the bottom loop by approaching the origin from the positive imaginary axis (or an angle of -270o ).

6072

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

The magnitude of the Nyquist plot as it crosses the negative real axis is 0.00174. It will not encircle the 1=K point until K = 1/0.00174 = 576.

i. 0 < K < 576 N = 0; P = 0 =) Z = N + P = 0 The closed-loop system is stable. ii. K > 576 N = 2; P = 0 =) Z = N + P = 2

6073 The closed-loop system has two unstable roots as veri…ed by the root locus. (c) The Bode plot shows an initial phase of 0o hence the Nyquist starts on the positive real axis at A’. The Bode ends with a phase of 180o hence the Nyquist ends the bottom loop by approaching the origin from the negative real axis (or an angle of -180o ).

It will never encircle the -1/K point, hence it is always stable. The root locus below con…rms that.

6074

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

N = 0; P = 0 =) Z = N + P = 0 The closed-loop system is stable for any K > 0: 20. Draw a Nyquist plot for

KG(s) =

K(s + 1) s(s + 3)

(1)

choosing the contour to be to the right of the singularity on the j!-axis. and determine the range of K for which the system is stable using the Nyquist Criterion. Then redo the Nyquist plot, this time choosing the contour to be to the left of the singularity on the imaginary axis and again check the range of K for which the system is stable using the Nyquist Criterion. Are the answers the same? Should they be? Solution : If you choose the contour to the right of the singularity on the origin, the Nyquist plot looks like this : 1 From the Nyquist plot, the range of K for stability is K < 0 (N = 0; P = 0 =) Z = N + P = 0): So the system is stable for K > 0:

Similarly, in the case with the contour to the left of the singularity on the origin, the Nyquist plot is:

6075

Figure 6.89: Control system for Problem 21

1 < 0 (N = From the Nyquist plot, the range of K for stability is K 1; P = 1 =) Z = N + P = 0): So the system is stable for K > 0:

The way of choosing the contour around singularity on the j!-axis does not a¤ect its stability criterion. The results should be the same in either way. However, it is somewhat less cumbersome to pick the contour to the right of a pole on the imaginary axis so that there are no unstable poles within the contour, hence P=0. 21. Draw the Nyquist plot for the system in Fig. 6.89. Using the Nyquist stability criterion, determine the range of K for which the system is stable. Consider both positive and negative values of K. Solution : The characteristic equation: 1+K

1 1 =0 (s2 + 2s + 2) (s + 1)

G(s) =

(s +

1 + 2s + 2)

1)(s2

6076

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

For positive K, note that the magnitude of the Nyquist plot as it crosses the negative real axis is 0.1, hence K < 10 for stability. For negative K, the entire Nyquist plot is essentially ‡ipped about the imaginary axis, thus the magnitude where it crosses the negative real axis will be 0.5 and the stability limit is that jKj < 2 Therefore, the range of K for stability is 2 < K < 10: 22. (a) For ! = 0:1 to 100 rad/sec, sketch the phase of the minimum-phase system s+1 G(s) = s + 10 s=j! and the nonminimum-phase system G(s) = noting that \(j!

s 1 s + 10

; s=j!

1) decreases with ! rather than increasing.

(b) Does a RHP zero a¤ect the relationship between the 1 encirclements on a polar plot and the number of unstable closed-loop roots in Eq. (6.28)? (c) Sketch the phase of the following unstable system for ! = 0:1 to 100 rad/sec: s+1 G(s) = : s 10 s=j! (d) Check the stability of the systems in (a) and (c) using the Nyquist criterion on KG(s). Determine the range of K for which the closedloop system is stable, and check your results qualitatively using a rough root-locus sketch.

6077 Solution :

(a) Minimum phase system,

G1 (j!) =

s+1 js=j! s + 10

Bode plot for Prob. 6.22 0

Magnitude

10

10

-1

-1

10

10

0

10

1

2

10

ω (rad/s ec) 100

Phase (deg)

80 60 40 20 0 -1 10

10

0

10 ω (rad/s ec)

Non-minimum phase system,

G2 (j!) =

s 1 js=j! s + 10

1

2

10

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Bode plot for Prob. 6.22 0

Magnitude

10

10

-1

-1

10

10

0

10

1

2

10

ω (rad/s ec) 0

Phase (deg)

6078

-50 -100 -150 -200 -1 10

10

0

10

1

ω (rad/s ec)

(b) No, a RHP zero doesn’t a¤ect the relationship between the 1 encirclements on the Nyquist plot and the number of unstable closed-loop roots in Eq. (6.28).

(c) Unstable system:

G3 (j!) =

s+1 js=j! s 10

2

10

6079 Bode plot for Prob. 6.22 (c) 0

Magnitude

10

10

-1

-1

10

10

0

10

1

2

10

ω (rad/s ec)

Phase (deg)

0 -50 -100 -150 -200 -1 10

10

0

10

1

ω (rad/s ec)

i. Minimum phase system G1 (j!): For any K > 0; N = 0; P = 0 =) Z = 0 =) The system is stable, as veri…ed by the root locus being entirely in the LHP.

2

10

6080

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

ii. Non-minimum phase system G2 (j!): the be encircled if K < 1:

1=K point will not

0 < K < 1 N = 0; P = 0 =) Z = 0 =) Stable 1 10; however, P = 1, so

1=K point will be encircled if

0 < K < 10 : N = 0; P = 1 =) Z = 1 =) Unstable 10 < K : N = 1; P = 1 =) Z = 0 =) Stable This is veri…ed by the Root Locus shown below right, where the locus crosses the imaginary axis when K = 10, and stays in the LHP for K > 10:

23. Nyquist plots and their classical plane curves: Determine the Nyquist plot using Matlab for the systems given below with K = 1 and verify that

6082

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD the beginning point and end point for the j! > 0 portion have the correct magnitude and phase: (a) the classical curve called Cayley’s Sextic, discovered by Maclaurin in 1718 1 KG(s) = K (s + 1)3 (b) the classical curve called the Cissoid, meaning ivy-shaped KG(s) = K

1 s(s + 1)

(c) the classical curve called the Folium of Kepler, studied by Kepler in 1609. 1 KG(s) = K (s 1)(s + 1)2 (d) the classical curve called the Folium (not Kepler’s) KG(s) = K

(s

1 1)(s + 2)

(e) the classical curve called the Nephroid, meaning kidney-shaped. KG(s) = K

2(s + 1)(s2 4s + 1) (s 1)3

(f) the classical curve called Nephroid of Freeth, named after the English mathematician T. J. Freeth. KG(s) = K

(s + 1)(s2 + 3) 4(s 1)3

(g) a shifted Nephroid of Freeth KG(s) = K

(s2 + 1) (s 1)3

Solution : These are all accomplished by using Matlab’s Nyquist function. All interesting shapes. To check the magnitude and phase for each, plug in s = 0 and s = inf and then compare those values with the beginning and end points on the Nyquist diagrams. (a)

6083

Nyquist Diagram 0.8

0.6

Imaginary Axis

0.4

0.2

0

-0.2

-0.4

-0.6

-0.8 -1

-0.8

-0.6

-0.4

-0.2

0 Real Axis

0.2

0.4

0.6

0.8

1

-0.4

-0.3

-0.2

-0.1

0

(b)

Nyquist Diagram 20

15

Imaginary Axis

10

5

0

-5

-10

-15

-20 -1

(c)

-0.9

-0.8

-0.7

-0.6

-0.5 Real Axis

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Nyquist Diagram 0.4

0.3

Imaginary Axis

0.2

0.1

0

-0.1

-0.2

-0.3

-0.4 -1

-0.9

-0.8

-0.7

-0.6

-0.5 Real Axis

-0.4

-0.3

-0.2

-0.1

0

-0.4

-0.3

-0.2

-0.1

0

(d)

Nyquist Diagram 0.2

0.15

0.1

Imaginary Axis

6084

0.05

0

-0.05

-0.1

-0.15

-0.2 -1

(e)

-0.9

-0.8

-0.7

-0.6

-0.5 Real Axis

6085

Nyquist Diagram 4

3

Imaginary Axis

2

1

0

-1

-2

-3

-4 -3

-2

-1

0 Real Axis

1

2

3

(f)

Nyquist Diagram 0.8

0.6

Imaginary Axis

0.4

0.2

0

-0.2

-0.4

-0.6

-0.8 -1

(g)

-0.8

-0.6

-0.4 -0.2 Real Axis

0

0.2

0.4

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Nyquist Diagram 0.2 0.15 0.1 0.05 Imaginary Axis

6086

0 -0.05 -0.1 -0.15 -0.2 -0.25 -0.4

-0.3

-0.2

-0.1 Real Axis

0

0.1

6087

Figure 6.90: Nyquist plot for Problem 24

(a)

Problems and Solutions for Section 6.4

24. The Nyquist plot for some actual control systems resembles the one shown in Fig.6.90. What are the gain and phase margin(s) for the system of Fig. 6.90 given that = 0:4; = 1:3; and = 40o : Describe what happens to the stability of the system as the gain goes from zero to a very large value. Sketch what the corresponding root locus must look like for such a system. Also sketch what the corresponding Bode plots would look like for the system. Solution : The phase margin is de…ned as in Figure 6.34, P M = (! = ! ), but now there are several gain margins! If the system gain is increased (multiplied) by j 1 j or decreased (divided) by j j, then the system will go unstable. This is a conditionally stable system. See Figure 6.40 for a typical root locus of a conditionally stable system.

gain margin gain margin

= =

-20 log j jdB (! = ! H ) +20 log j jdB (! = ! L )

For a conditionally stable type of system as in Fig. 6.40, the Bode phase plot crosses -180 twice; however, for this problem we see from the Nyquist plot that it crosses 3 times! For very low values of gain, the entire

6088

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Nyquist plot would be shrunk, and the -1 point would occur to the left of the negative real axis crossing at ! o , so there would be no encirclements and the system would be stable. As the gain increases, the -1 point occurs between ! o and ! L so there is an encirclement and the system is unstable. Further increase of the gain causes the -1 point to occur between ! L and ! H (as shown in Fig. 6.90) so there is no encirclement and the system is stable. Even more increase in the gain would cause the -1 point to occur between ! H and the origin where there is an encirclement and the system is unstable. The root locus would look like Fig. 6.40 except that the very low gain portion of the loci would start in the LHP before they loop out into the RHP as in Fig. 6.40. The Bode plot would be vaguely like that drawn below:

25. The Bode plot for G(s) =

100[(s=10) + 1] s[(s=1) 1][(s=100) + 1]

is shown in Fig. 6.91. (a) Why does the phase start at -270o at the low frequencies? (b) Sketch the Nyquist plot for G(s). (c) Is the closed-loop system shown in Fig. 6.92 stable?

6089

Figure 6.91: Bode plot for Problem 25

(d) Will the system be stable if the gain is lowered by a factor of 100? Make a rough sketch of a root locus for the system and qualitatively con…rm your answer Solution : (a) From the root locus, the phase at the low frequencies (! = 0+) is calculated as : The phase at the point fs = j!(! = 0+)g = 180 (pole : s = 1) 90 (pole : s = 0) + 0 (zero : s = = 270

10) + 0 (pole : s =

Or, more simply, the RHP pole at s = +1 causes a 180o shift from the 90o that you would expect from a normal system with all the singularities in the LHP.

100)

6090

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Figure 6.92: Control system for Problem 26

(b) The Nyquist plot for G(s) :

(c) As the Nyquist shows, there is one counter-clockwise encirclement of -1. =) N =

1

We have one pole in RHP =) P = 1 Z = N + P = 1 + 1 = 0 =) The closed-loop system is stable. (d) The system goes unstable if the gain is lowered by a factor of 100. 26. Suppose that in Fig. 6.92, G(s) =

25(s + 1) : s(s + 2)(s2 + 2s + 16)

Use MATLAB’s margin to calculate the PM and GM for G(s) and, based on the Bode plots, conclude which margin would provide more useful information to the control designer for this system. Solution :

6091

Bode Diagram Gm = 3.91 dB (at 4.22 rad/sec) , Pm = 101 deg (at 1.08 rad/sec) 20

Magnitude (dB)

0 -20 -40 -60 -80 -100 -45

Phase (deg)

-90 -135 -180 -225 -270 10

-1

10

0

10

1

Frequency (rad/sec)

From the Bode plot, P M = 101 deg; GM = 3:9db = 1:57 Since both PM and GM are positive, we can say that the closed-loop of this system is stable. But GM is so small that we must be careful not to increase the gain much, which leads the closed-loop system to be unstable. Clearly, the GM is the more important margin for this example. 27. Consider the system given in Fig. 6.93. (a) Use MATLAB to obtain Bode plots for K = 1 and use the plots to estimate the range of K for which the system will be stable. (b) Verify the stable range of K by using margin to determine PM for selected values of K. (c) Use rlocus and rloc…nd to determine the values of K at the stability boundaries.

10

2

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Figure 6.93: Control system for Problem 27

(d) Sketch the Nyquist plot of the system, and use it to verify the number of unstable roots for the unstable ranges of K. (e) Using Routh’s criterion, determine the ranges of K for closed-loop stability of this system. Solution : (a) The Bode plot for K = 1 is :

Magnitude

10

10

10

Bode plot for Prob. 6.27 (a): K =1

0

-1

-2 -1

10

0

10 Frequency (rad/s ec)

1

10

-160

Phase (deg)

6092

-170 -180 -190 -200 -1 10

0

10 ω (rad/s ec)

1

10

6093 From the Bode plot, the closed-loop system is unstable for K = 1. But we can make the closed-system stable with positive GM by increasing the gain K up to the crossover frequency reaches at ! = 1:414 rad/sec (K = 2), where the phase plot crosses the 180 line. Therefore :

1 < K < 2 =) The closed-loop system is stable.

(b) For example, P M = 6:66 deg for K = 1:5.

Bode Diagram Gm = 2.5 dB (at 1.41 rad/sec) , Pm = 6.66 deg (at 1.08 rad/sec) 20

Magnitude (dB)

0 -20 -40 -60 -80 -160

Phase (deg)

-170

-180

-190

-200 10

-1

10

0

10 Frequency (rad/sec)

(c) Root locus is :

1

10

2

6094

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

j!-crossing :

1+K

j! + 2 (j!)3 + (j!)2

2

=0

! 2 2K + 2 = 0 !(! 2 K) = 0

K = 2; ! =

p

2; or K = 1; ! = 0

Therefore,

1 < K < 2 =) The closed-loop system is stable.

6095

(d) i. 0 < K < 1 N = 0; P = 1 =) Z = 1 One unstable closed-loop root. ii. 1 < K < 2 N = 1; P = 1 =) Z = 0 Stable. iii. 2 < K N = 1; P = 1 =) Z = 2 Two unstable closed-loop roots. (e) The closed-loop transfer function of this system is : y(s) r(s)

k =

1 s

1

s+2 s 1 (s + 1)2 + 1 K(s2 + 2s + 2) s3 + s2 + Ks + 2K 2 1+k

=

1

So the characteristic equation is : =) s3 + s2 + Ks + 2K

2=0

Using the Routh’s criterion, s3 s2 s1 s0 For stability,

: : : :

1 1 2 K 2k = 2

K 2K 2 0

6096

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

2 K > 0 2K 2 > 0 =) 2 > K > 1 0 < K < 1 Unstable 1 40 . Use Matlab to verify and/or re…ne your design so that it meets the speci…cations. Solution : Use a lead compensation : D(s) =

Ts + 1 ; Ts + 1

From the speci…cation, Kv = 20 sec

1

>1

;

=)

Kv = lim sD(s)G(s) = K = 20

=)

K = 20

s!0

3

10

6142

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD From a hand sketch of the uncompensated Bode plot asymptotes, we see that the slope at crossover is -2, hence the PM will be poor. In fact, an exact computation shows that P M = 12:75 (at ! c = 4:42 rad/sec) Adding a lead compensation s +1 3 D(s) = s +1 30 will provide a -1 slope in the vicinity of crossover and should provide plenty of PM. The Bode plot below veri…es that indeed it did and shows that the P M = 62 at a crossover frequency = 7 rad/sec thus meeting all specs.

Bode Diagram Gm = Inf dB (at Inf rad/sec) , Pm = 61.8 deg (at 6.99 rad/sec) 100

Magnitude (dB)

50

0

-50

Phase (deg)

-100 -90

-135

-180 10

-2

10

-1

0

10 10 Frequency (rad/sec)

1

54. Consider a satellite-attitude control system with the transfer function G(s) =

0:05(s + 25) : s2 (s2 + 0:1s + 4)

10

2

10

3

6143 Amplitude-stabilize the system using lead compensation so that GM 2 (6 db), and PM 45 , keeping the bandwidth as high as possible with a single lead.

Solution :

The sketch of the uncompensated Bode plot asymptotes shows that the slope at crossover is -2; therefore, a lead compensator will be required in order to have a hope of meeting the PM requirement. Furthermore, the resonant peak needs to be kept below magnitude 1 so that it has no chance of causing an instability (this is amplitude stabilization). This latter requirement means we must lower gain at the resonance. Using the single lead compensator,

D(s) =

(s + 0:06) (s + 6)

will lower the low frequency gain by a factor of 100, provide a -1 slope at crossover, and will lower the gain some at the resonance. Thus it is a good …rst cut at a compensation. The Matlab Bode plot shows the uncompensated and compensated and veri…es our intent. Note especially that the resonant peak never crosses magnitude 1 for the compensated (dashed) case.

6144

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Bode Diagram s

5

Magnitude

10

0

10

-5

10

-2

10

10

-1

0

10 ω (rad/s ec)

1

10

2

10

Phase (deg)

-100

-200 -300 -400 -2 10

10

-1

0

10 ω (rad/s ec)

1

10

The Matlab margin routine shows a GM = 6:3 db and P M = 48 thus meeting all specs. 55. In one mode of operation the autopilot of a jet transport is used to control altitude. For the purpose of designing the altitude portion of the autopilot loop, only the long-period airplane dynamics are important. The linearized relationship between altitude and elevator angle for the longperiod dynamics is G(s) =

h(s) 20(s + 0:01) ft = : (s) s(s2 + 0:01s + 0:0025) deg

The autopilot receives from the altimeter an electrical signal proportional to altitude. This signal is compared with a command signal (proportional to the altitude selected by the pilot), and the di¤erence provides an error signal. The error signal is processed through compensation, and the result is used to command the elevator actuators. A block diagram of this system is shown in Fig. 6.103. You have been given the task of designing the compensation. Begin by considering a proportional control law D(s) = K.

2

10

6145

Figure 6.103: Control system for Problem 55

(a) Use Matlab to draw a Bode plot of the open-loop system for D(s) = K = 1. (b) What value of K would provide a crossover frequency (i.e., where jGj = 1) of 0.16 rad/sec? (c) For this value of K, would the system be stable if the loop were closed?

(d) What is the PM for this value of K? (e) Sketch the Nyquist plot of the system, and locate carefully any points where the phase angle is 180 or the magnitude is unity. (f) Use Matlab to plot the root locus with respect to K, and locate the roots for your value of K from part (b). (g) What steady-state error would result if the command was a step change in altitude of 1000 ft? For parts (h)and (i), assume a compensator of the form D(s) = K

Ts + 1 : Ts + 1

(h) Choose the parameters K, T , and so that the crossover frequency is 0.16 rad/sec and the PM is greater that 50 . Verify your design by superimposing a Bode plot of D(s)G(s)=K on top of the Bode plot you obtained for part (a), and measure the PM directly. (i) Use Matlab to plot the root locus with respect to K for the system including the compensator you designed in part (h). Locate the roots for your value of K from part (h). (j) Altitude autopilots also have a mode where the rate of climb is sensed directly and commanded by the pilot. i. Sketch the block diagram for this mode, ii. de…ne the pertinent G(s); iii. design D(s) so that the system has the same crossover frequency as the altitude hold mode and the PM is greater than 50 Solution :

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD The plant transfer function : h(s) = (s)

s +1 0:01 2 0:1 s +2 s+1 0:05 0:05 80

s

(a) See the Bode plot :

Bode Diagram Gm = Inf dB (at Inf rad/sec) , Pm = 0.386 deg (at 0.16 rad/sec) 80

Magnitude (dB)

60 40 20 0 -20 -40 0

-45 Phase (deg)

6146

-90

-135

-180 10

-4

10

-3

(b) Since jGj = 865 at ! = 0:16, K=

1 j!=0:16 = 0:0012 jGj

(c) The system would be stable, but poorly damped. (d) P M = 0:39 (e) The Nyquist plot for D(j!)G(j!) :

-2

10 Frequency (rad/sec)

10

-1

10

0

6147

The phase angle never quite reaches

180 .

(f) See the Root locus :

The closed-loop roots for K = 0:0012 are : s=

0:009;

0:005

j0:16

6148

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD (g) The steady-state error e1 : ess

=

lim s

s!0

=

1 h(s) 1+K (s)

1000 s

0

as it should be for this Type 1 system. (h) Phase margin of the plant : P M = 0:39 (! c = 0:16 rad/sec) Necessary phase lead and

1

:

necessary phase lead = 50

0:39

' 50

From Fig. 6.54 : =)

1

=8

Set the maximum phase lead frequency at ! c : !=p

1 T

= ! c = 0:16 =) T = 18

so the compensation is D(s) = K

18s + 1 2:2s + 1

For a gain K, we want jD(j! c )G(j! c )j = 1 at ! = ! c = 0:16: So evaluate via Matlab D(j! c )G(j! c ) K

and …nd it

=

2:5

103

! c =0:16

=) K = Therefore the compensation is : D(s) = 4:0

10

4

18s + 1 2:2s + 1

which results in the Phase margin : P M = 52 (! c = 0:16 rad/sec)

1 2:5

103

= 4:0

10

4

6149

Bode Diagram s

6

Magnitude

10

4

10

2

10

10

-3

10

-2

-1

10

0

10

ω (rad/s ec)

Phase (deg)

0 -50 -100 -150 -200 -3 10

10

-2

-1

10 ω (rad/s ec)

(i) See the Root locus :

0

10

6150

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

The closed-loop roots for K = 4:0 s=

0:27;

0:0074;

10

4

are :

0:095

j0:099

(j) In this case, the reference input and the feedback parameter are the rate of climb. i. The block diagram for this mode is :

ii. De…ne G(s) as : s +1 0:01 s 2 0:1 s +2 s+1 0:05 0:05 iii. By evaluating the gain of G(s) at ! = ! c = 0:16; and setting K equal to its inverse, we see that proportional feedback : _ h(s) G(s) = = (s)

80

D(s) = K = 0:0072 satis…es the given speci…cations by providing: P M = 90 (! c = 0:16 rad/sec) The Bode plot of the compensated system is :

6151

Bode Diagram Gm = Inf , Pm = 90.4 deg (at 0.16 rad/sec) 30

Magnitude (dB)

20 10 0 -10 -20 90

Phase (deg)

45

0

-45

-90 10

-4

10

-3

-2

10 Frequency (rad/sec)

56. For a system with open-loop transfer function c 10 ; G(s) = s[(s=1:4) + 1][(s=3) + 1] design a lag compensator with unity DC gain so that PM the approximate bandwidth of this system?

40 . What is

Solution : Lag compensation design : Use D(s) =

Ts + 1 Ts + 1

K=1 so that DC gain of D(s) = 1. (a) Find the stability margins of the plant without compensation by plotting the Bode, …nd that: PM GM

= =

20 (! c = 3:0 rad/sec) 0:44 (! = 2:05 rad/sec)

10

-1

10

0

6152

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD (b) The lag compensation needs to lower the crossover frequency so that a P M ' 40 will result, so we see from the uncompensated Bode that we need the crossover at about =) ! c;new = 0:81 where jG(j! c )j = 10:4 so the lag needs to lower the gain at ! c;new from 10.4 to 1. (c) Pick the zero breakpoint of the lag to avoid in‡uencing the phase at ! = ! c;new by picking it a factor of 20 below the crossover, so ! c;new 1 = T 20 =) T = 25 (d) Choose

:

Since D(j!) =

1

for ! >> 1

1 T

, let

=

1 jG(j! c;new )j

= jG(j! c;new )j = 10:4 (e) Compensation : D(s) =

s 0:04 + 1 s 0:0038 + 1

(f) Stability margins of the compensated system : PM GM

= =

42 (! c = 0:8 rad/sec) 4:4 (! = 2:0 rad/sec)

Approximate bandwidth ! BW : P M = 42

=) ! BW = 2! c = 1:6 (rad/sec)

6153

Bode plot

4

Magnitude

10

Uncom pens ated

2

10

0

10

Com pens ated -2

10

10

-3

10

-2

-1

10 ω (rad/s ec)

0

10

1

10

Phase (deg)

-100 -150 -200 -250 10

-3

10

-2

-1

10 ω (rad/s ec)

57. For the ship-steering system in Problem 39, (a) Design a compensator that meets the following speci…cations: i. velocity constant Kv = 2, ii. PM 50 , iii. unconditional stability (PM > 0 for all ! frequency).

! c , the crossover

(b) For your …nal design, draw a root locus with respect to K, and indicate the location of the closed-loop poles. Solution : The transfer function of the ship steering is V (s) K[ (s=0:142) + 1] = G(s) = : s(s=0:325 + 1)(s=0:0362) + 1) r (s) (a) Since the velocity constant, Kv must be 2, we require that K = 2:

0

10

1

10

6154

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD i. The phase margin of the uncompensated ship is PM =

111 (! c = 0:363 rad/sec)

which means it would be impossible to stabilize this system with one lead compensation, since the maximum phase increase would be 90o : There is no speci…cation leading to maintaining a high bandwidth, so the use of lag compensation appears to be the best choice. So we use a lag compensation: D(s) =

Ts + 1 Ts + 1

ii. The crossover frequency which provides P M ' 50 is obtained by looking at the uncompensated Bode plot below, where we see that the crossover frequency needs to be lowered to ! c;new = 0:017; where the uncompensated gain is jG(j! c;new )j = 107 iii. Keep the zero of the lag a factor of 20 below the crossover to keep the phase lag from the compensation from fouling up the PM, so we …nd: 1 T

= =)

iv. Choose

! c;new 20 T = 1:2

103

so that the gain reduction is achieved at crossover : = jG(j! c;new )j = 107

1 ) T v. So the compensation is : (D(j!) '

1

for !

D(s) =

1200s + 1 = 12:6s + 1

s 0:0008 + 1 s 0:08 + 1

vi. Stability margins of the compensated system : PM GM

= 52:1 (! c = 0:017 rad/sec) = 5:32 (! = 0:057 rad/sec)

and the system is unconditionally stable since the phase > 0 for all ! < ! c :as can be seen by the plot below.

6155 (b) See the root locus. (Note that this is a zero degree root locus.) The closed-loop roots for K = 2 are :

s=

0:33;

0:0009;

0:014

j0:021

Bode Diagram s 5

Magnitude

10

Uncom pens ated

0

10

10

-5

-4

10

-3

-2

-3

-2

10 10 Frequency (rad/s ec)

-1

10

0

10

0

Phase (deg)

Uncom pens ated -100 -200 -300 -400 -5 10

-4

10

10 10 Frequency (rad/s ec)

-1

10

0

10

6156

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

58. For a unity feedback system with G(s) =

1 s s( 20

+

s2 1)( 100 2

s + 0:5 100 + 1)

(2)

(a) A lead compensator is introduced with = 1=5 and a zero at 1=T = 20. How must the gain be changed to obtain crossover at ! c = 31:6 rad/sec, and what is the resulting value of Kv ? (b) With the lead compensator in place, what is the required value of K for a lag compensator that will readjust the gain to a Kv value of 100? (c) Place the pole of the lag compensator at 3.16 rad/sec, and determine the zero location that will maintain the crossover frequency at ! c = 31:6 rad/sec. Plot the compensated frequency response on the same graph. (d) Determine the PM of the compensated design. Solution : (a) From a sketch of the asymptotes with the lead compensation (with K1 = 1) : s +1 D1 (s) = K1 20 s 100 + 1 in place, we see that the slope is -1 from zero frequency to ! = 100 rad/sec. Therefore, to obtain crossover at ! c = 31:6 rad/sec, the gain K1 = 31:6 is required. Therefore, Kv = 31:6

6157 (b) To increase Kv to be 100, we need an additional gain of 3.16 from the lag compensator at very low frequencies to yield Kv = 100: (c) For a low frequency gain increase of 3.16, and the pole at 3.16 rad/sec, the zero needs to be at 10 in order to maintain the crossover at ! c = 31:6 rad/sec. So the lag compensator is s +1 10 D2 (s) = 3:16 s +1 3:16 and D1 (s)D2 (s) = 100

s 20 + 1 s 100 + 1

s +1 10 s +1 3:16

The Bode plots of the system before and after adding the lag compensation are

Bode Diagram s

2

10

Magnitude

Lead and Lag Lead only 0

10

-2

10

10

0

10

1

2

10

3

10

ω (rad/s ec)

Phase (deg)

0 -100 -200 -300 -400 0 10

10

1

2

10 ω (rad/s ec)

3

10

6158

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD (d) By using the margin routine from Matlab, we see that P M = 49 (! c = 34:5 deg/sec)

59. Golden Nugget Airlines had great success with their free bar near the tail of the airplane. (See Problem 5.39) However, when they purchased a much larger airplane to handle the passenger demand, they discovered that there was some ‡exibility in the fuselage that caused a lot of unpleasant yawing motion at the rear of the airplane when in turbulence and was causing the revelers to spill their drinks. The approximate transfer function for the dutch roll mode (See Section 10.3.1) is 8:75(4s2 + 0:4s + 1) r(s) = (s=0:01 + 1)(s2 + 0:24s + 1) r (s) where r is the airplane’s yaw rate and r is the rudder angle. In performing a Finite Element Analysis (FEA) of the fuselage structure and adding those dynamics to the dutch roll motion, they found that the transfer function needed additional terms that re‡ected the fuselage lateral bending that occurred due to excitation from the rudder and turbulence. The revised transfer function is r(s) 8:75(4s2 + 0:4s + 1) 1 = 2 2 s (s=0:01 + 1)(s + 0:24s + 1) ( !2 + 2 !s + 1) r (s) b b

where ! b is the frequency of the bending mode (= 10 rad/sec) and is the bending mode damping ratio (= 0:02). Most swept wing airplanes have a “yaw damper”which essentially feeds back yaw rate measured by a rate gyro to the rudder with a simple proportional control law. For the new Golden Nugget airplane, the proportional feedback gain, K = 1; where r (s)

=

Kr(s):

(3)

(a) Make a Bode plot of the open-loop system, determine the PM and GM for the nominal design, and plot the step response and Bode magnitude of the closed-loop system. What is the frequency of the lightly damped mode that is causing the di¢ culty? (b) Investigate remedies to quiet down the oscillations, but maintain the same low frequency gain in order not to a¤ect the quality of the dutch roll damping provided by the yaw rate feedback. Speci…cally, investigate one at a time: i. increasing the damping of the bending mode from = 0:02 to = 0:04: (Would require adding energy absorbing material in the fuselage structure) ii. increasing the frequency of the bending mode from ! b = 10 rad/sec to ! b = 20 rad/sec. (Would require stronger and heavier structural elements)

6159 iii. adding a low pass …lter in the feedback, that is, replace K in Eq. (3) with KD(s) where

D(s) =

1 s=

p

+1

:

(4)

Pick p so that the objectionable features of the bending mode are reduced while maintaining the PM 60o : iv. adding a notch …lter as described in Section 5.4.3. Pick the frequency of the notch zero to be at ! b with a damping of = 0:04 and pick the denominator poles to be (s=100 + 1)2 keeping the DC gain of the …lter = 1.

(c) Investigate the sensitivity of the two compensated designs above (iii and iv) by determining the e¤ect of a reduction in the bending mode frequency of -10%. Speci…cally, re-examine the two designs by tabulating the GM, PM, closed loop bending mode damping ratio and resonant peak amplitude, and qualitatively describe the di¤erences in the step response.

(d) What do you recommend to Golden Nugget to help their customers quit spilling their drinks? (Telling them to get back in their seats is not an acceptable answer for this problem! Make the recommendation in terms of improvements to the yaw damper.)

Solution :

(a) The Bode plot of the open-loop system is :

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Bode Diagram Gm = 1.28 dB (at 10 rad/sec) , Pm = 97.6 deg (at 0.0833 rad/sec) 20

Magnitude (dB)

0 -20 -40 -60 -80 -100 90

0 Phase (deg)

6160

-90

-180

-270 10

-4

PM GM

10

-3

10

-2

-1

10 Frequency (rad/sec)

10

0

= 97:6 (! = 0:0833 rad/sec) = 1:28 (! = 10:0 rad/sec)

The low GM is caused by the resonance being close to instability. The closed-loop system unit step response is :

10

1

10

2

6161 Unit Step Res pons e 0.9 0.8 0.7

Amplitude

0.6 0.5 0.4 0.3 0.2 0.1 0

0

10

20

30

The Bode plot of the closed-loop system is :

40 50 Tim e (s ec)

60

70

80

90

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Bode plot 0

Magnitude

10

-5

10

-3

10

-2

10

10

-1

10 ω (rad/s ec)

0

1

10

2

10

0 Phase (deg)

6162

-100 -200 -300 -3 10

-2

10

10

-1

10 ω (rad/s ec)

0

From the Bode plot of the closed -loop system, the frequency of the lightly damped mode is :

! ' 10 rad/sec

and this is borne out by the step response that shows a lightly damped oscillation at 1.6 Hz or 10 rad/sec.

i. The Bode plot of the system with the bending mode damping increased from = 0:02 to = 0:04 is :

1

10

2

10

6163 Bode Diagram Gm = 7.31 dB (at 10 rad/sec) , Pm = 97.6 deg (at 0.0833 rad/sec) 20

Magnitude (dB)

0 -20 -40 -60 -80 -100 90

Phase (deg)

0

-90

-180

-270 10

-4

PM GM

10

-3

10

-2

-1

10 Frequency (rad/sec)

10

= 97:6 (! = 0:0833 rad/sec) = 7:31 (! = 10:0 rad/sec)

and we see that the GM has increased considerably because the resonant peak is well below magnitude 1; thus the system will be much better behaved. ii. The Bode plot of this system (! b = 10 rad/sec =) ! b = 20 rad/sec) is :

0

10

1

10

2

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Bode D iagram Gm = 7.35 dB (at 20 rad/sec) , Pm = 97.6 deg (at 0.0833 rad/sec ) 20

Magnitude (dB)

0 -20 -40 -60 -80 90

0 Phase (deg)

6164

-90

-180

-270 10

-4

10

PM GM

-3

10

-2

-1

10 10 Frequency (rad/sec)

0

= 97:6 (! = 0:0833 rad/sec) = 7:34 (! = 20:0 rad/sec)

and again, the GM is much improved and the resonant peak is signi…cantly reduced from magnitude 1.

iii. By picking up pass …lter is :

p

= 1; the Bode plot of the system with the low

10

1

2

10

6165 Bode Diagram Gm = 35 dB (at 8.62 rad/sec) , Pm = 92.9 deg (at 0.0831 rad/sec) 50

Magnitude (dB)

0

-50

-100

-150 90

Phase (deg)

0 -90 -180 -270 -360 10

-4

PM GM

10

-3

10

-2

-1

10 Frequency (rad/sec)

10

= 92:9 (! = 0:0831 rad/sec) = 34:97 (! = 8:62 rad/sec)

which are healthy margins and the resonant peak is, again, well below magnitude 1. iv. The Bode plot of the system with the given notch …lter is : PM GM

= 97:6 (! = 0:0833 rad/sec) = 55:3 (! = 99:7 rad/sec)

which are the healthiest margins of all the designs since the notch …lter has essentially canceled the bending mode resonant peak. (b) Generally, the notch …lter is very sensitive to where to place the notch zeros in order to reduce the lightly damped resonant peak. So if you want to use the notch …lter, you must have a good estimation of the location of the bending mode poles and the poles must remain at that location for all aircraft conditions. On the other hand, the low pass …lter is relatively robust to where to place its break point.

0

10

1

10

2

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Bode Diagram Gm = 55.3 dB (at 99.7 rad/sec) , Pm = 97.6 deg (at 0.0833 rad/sec) 50

Magnitude (dB)

0

-50

-100

-150 90

0 Phase (deg)

6166

-90

-180

-270 10

-4

10

-3

10

-2

-1

0

10 10 Frequency (rad/sec)

10

1

10

2

10

3

6167 Evaluation of the margins with the bending mode frequency lowered by 10% will show a drastic reduction in the margins for the notch …lter and very little reduction for the low pass …lter.

GM PM Closed-loop bending mode damping ratio Resonant peak

Low Pass Filter 34:97 (! = 8:62 rad/sec) 92:9 (! = 0:0831 rad/sec)

Notch Filter 55:3 (! = 99:7 rad/sec) 97:6 (! = 0:0833 rad/sec)

' 0:02

' 0:04

0.087

0.068

The magnitude plots of the closed-loop systems are :

Magnitude (Notch Filter)

Magnitude (Low Pass Filter)

Bode plot 0

10

-5

10

-3

10

-2

10

10

-1

10 ω (rad/s ec)

0

1

10

2

10

0

10

-5

10

-3

10

-2

10

10

-1

10 ω (rad/s ec)

0

The closed-loop step responses are : (c) While increasing the natural damping of the system would be the best solution, it might be di¢ cult and expensive to carry out. Likewise, increasing the frequency typically is expensive and makes the structure heavier, not a good idea in an aircraft. Of the remaining

1

10

2

10

6168

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Unit Step Res pons e 0.9 low pas s filter 0.8

notch filter

0.7

Amplitude

0.6 0.5 0.4 0.3 0.2 0.1 0

0

10

20

30 Tim e (s ec)

40

50

60

6169 two options, it is a better design to use a low pass …lter because of its reduced sensitivity to mismatches in the bending mode frequency. Therefore, the best recommendation would be to use the low pass …lter. 60. Consider a system with the open-loop transfer function (loop gain) G(s) =

1 : s(s + 1)(s=10 + 1)

(a) Create the Bode plot for the system, and …nd GM and PM. (b) Compute the sensitivity function and plot its magnitude frequency response. (c) Compute the Vector Margin (VM). Solution : (a) The Bode plot is :

Bode plot

Magnitude

0

10

-5

10

-2

10

10

-1

0

10 Frequency (rad/s ec)

1

10

2

10

-50

Phase (deg)

-100 -150 -200 -250 -300 -2 10

10

-1

0

10 Frequency (rad/s ec)

1

10

2

10

6170

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD (b) Sensitivity function is : S(s)

=

1 1 + G(s) 1

=

1

1+

s(s + 1)(

s + 1) 10

The magnitude frequency response of this sensitivity function is : Frequency Res pons e of the Sens itivity Function

0

Magnitude

10

-1

10

10

-1

10

0

1

10 ω (rad/s ec)

(c) Vector Margin is de…ned as : VM

1 js(j!)j

=

min

=

1 = 0:62 1:61

!

61. Prove that the sensitivity function S(s) has magnitude greater than 1 inside a circle with a radius of 1 centered at the 1 point. What does

2

10

6171 this imply about the shape of the Nyquist plot if closed-loop control is to outperform open-loop control at all frequencies? Solution : S(s) =

1 1 + G(s)

Inside the unit circle, j1 + G(s)j < 1 which implies jS(s)j > 1.

Outside the unit circle, j1 + G(s)j > 1 which implies jS(s)j < 1. On the unit circle, j1 + G(s)j = 1 which means jS(s)j = 1.

If the closed-loop control is going to outperform open-loop control then jS(s)j 1 for all s. This means that the Nyquist plot must lie outside the circle of radius one centered at 1. 62. Consider the system in Fig. 6.102 with the plant transfer function G(s) =

10 : s(s=10 + 1)

We wish to design a compensator D(s) that satis…es the following design speci…cations: (a)

i. Kv = 100, ii. PM 45 , iii. sinusoidal inputs of up to 1 rad/sec to be reproduced with error,

2%

6172

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Figure 6.104: Control system constraints for Problem 62

iv. sinusoidal inputs with a frequency of greater than 100 rad/sec to be attenuated at the output to 5% of their input value. (b) Create the Bode plot of G(s), choosing the open-loop gain so that Kv = 100. (c) Show that a su¢ cient condition for meeting the speci…cation on sinusoidal inputs is that the magnitude plot lies outside the shaded regions in Fig. 6.104. Recall that KG Y = R 1 + KG

and

E 1 = : R 1 + KG

(d) Explain why introducing a lead network alone cannot meet the design speci…cations. (e) Explain why a lag network alone cannot meet the design speci…cations. (f) Develop a full design using a lead-lag compensator that meets all the design speci…cations, without altering the previously chosen low frequency open-loop gain. Solution : (a) To satisfy the given velocity constant Kv , Kv

=

lim sKG(s) = 10K = 100

s!0

=) K = 10 (b) The Bode plot of G(s) with the open-loop gain K = 10 is :

6173

Magnitude

Bode Diagram s

0

10

10

-1

10

0

1

10 Frequency (rad/s ec)

2

10

3

10

-80

Phase (deg)

-100 -120 -140 -160 -180 -200 -1 10

10

0

1

10 ω (rad/s ec)

(c) From the 3rd speci…cation, E R

= =)

1 < 0:02 (2%) 1 + KG jKGj > 49 (at ! < 1 rad/sec)

From the 4th speci…cation, Y R

= =)

2

10

KG < 0:05 (5%) 1 + KG jKGj < 0:0526 (at ! > 100 rad/sec)

which agree with the …gure. (d) A lead compensator may provide a su¢ cient PM, but it increases the gain at high frequency so that it violates the speci…cation above. (e) A lag compensator could satisfy the PM speci…cation by lowering the crossover frequency, but it would violate the low frequency speci…cation, W1 :

3

10

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD (f) One possible lead-lag compensator is : s s +1 +1 8:52 4:47 D(s) = 100 s s +1 +1 22:36 0:568 which meets all the speci…cation : Kv PM jKGj jKGj

= 100 = 47:7 (at ! c = 12:9 rad/sec) = 50:45 (at ! = 1 rad/sec) > 49 = 0:032 (at ! = 100 rad/sec) < 0:0526

The Bode plot of the compensated open-loop system D(s)G(s) is :

Bode Diagram Gm = Inf dB (at Inf rad/sec) , Pm = 47.7 deg (at 12.9 rad/sec) 100

Magnitude (dB)

50

0

-50

-100 -90

Phase (deg)

6174

-135

-180 10

-2

10

-1

0

10 10 Frequency (rad/sec)

1

10

2

10

3

6175

Problems and Solutions for Section 6.8 63. Assume that the system G(s) =

e Td s ; s + 10

has a 0.2-sec time delay (Td = 0:2 sec). While maintaining a phase margin 40 , …nd the maximum possible bandwidth using the following: (a) One lead-compensator section D(s) = K

s+a ; s+b

where b=a = 100; (b) Two lead-compensator sections D(s) = K

s+a s+b

2

;

where b=a = 10. (c) Comment on the statement in the text about the limitations on the bandwidth imposed by a delay. Solution : (a) One lead section : With b=a = 100, the lead compensator can add the maximum phase lead : max

= =

1 ab 1 + ab 78:6 deg ( at ! = 10a rad/sec) sin

1

By trial and error, a good compensator is : K

=

1202; a = 15 =) Da (s) = 1202

PM

=

40 (at ! c = 11:1 rad/sec)

s + 15 s + 1500

The Bode plot is shown below. Note that the phase is adjusted for the time delay by subtracting !Td at each frequency point while there is no e¤ect on the magnitude. For reference, the …gures also include the case of proportional control, which results in : K = 13:3; P M = 40 (at ! c = 8:6 rad/sec)

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Bode plot

1

Magnitude

10

0

10

-1

10

10

0

1

10 ω (rad/s ec)

2

10

0

Phase (deg)

6176

-100 -200 -300 -400 0 10

1

10 ω (rad/s ec)

(b) Two lead sections : With b=a = 10, the lead compensator can add the maximum phase lead : max

1

1 1+

a b a b

=

sin

=

54:9 deg ( at ! =

p

10a rad/sec)

By trial and error, one of the possible compensators is : K

=

1359; a = 70 =) Db (s) = 1359

PM

=

40 (at ! c = 9:6 rad/sec)

The Bode plot is shown below.

(s + 70)2 (s + 700)2

2

10

6177 Bode plot

0.3

Magnitude

10

-0.1

10

-0.5

10

10

0

1

10 ω (rad/s ec)

2

10

0

Phase (deg)

-50 -100 -150 -200 -250 0 10

1

10 ω (rad/s ec)

(c) The statement in the text is that it should be di¢ cult to stabilize a system with time delay at crossover frequencies, ! c & 3=Td : This problem con…rms this limit, as the best crossover frequency achieved was ! c = 9:6 rad/sec whereas 3=Td = 15 rad/sec. Since the bandwidth is approximately twice the crossover frequency, the limitations imposed on the bandwidth by the time delay is veri…ed. 64. Determine the range of K for which the following systems are stable: 4s

(a) G(s) = K e s

s

e (b) G(s) = K s(s+2)

Solution : (a) G(j!) G(j!) = 2:54; when \ = K K range of stability : 0 < K <

180

1 2:54

2

10

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Bode plot

2

Magnitude

10

1

10

0

10

-1

10

-1

0

10

10 Frequency (rad/s ec)

0

Phase (deg)

6178

-100 -200 -300 -400 -1 10

0

10 ω (rad/s ec)

(b)

G(j!) 1 G(j!) ; when \ = = 0:409 = K 2:45 K

range of stability : 0 < K < 2:45

180

6179

Bode plot

1

10

0

Magnitude

10

-1

10

-2

10

10

-1

0

10 ω (rad/s ec)

1

10

Phase (deg)

0 -100 -200 -300 -400 -1 10

0

10 ω (rad/s ec)

65. In Chapter 5, we used various approximations for the time delay, one of which is the …rst order Padé e

Td s

= H1 (s) =

1 Td s=2 : 1 + Td s=2

Using frequency response methods, the exact time delay H2 (s) = e can be used. cations.

Td s

:

Plot the phase of H1 (s) and H2 (s) and discuss the impli-

Solution : The approximation H1 (j!) and the true phase H2 (j!) are compared in the plot below:

1

10

6180

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Tim e Delay Phas e Lag 0

Phase (deg)

H -100

H

1 2

-200 -300 -400 -1 10

0

10 Norm alized Frequency,ω T (rad) d

H1 (j!) closely approximates the correct phase of the delay (phase of H2 (s)) for !Td . and progressively worsens above that frequency. The 2 implication is that the H1 (s) approximation should not be trusted for crossover frequencies ! c & . Instead, one should use the exact phase 2Td for the time delay given by H2 (s): 66. Consider the heat exchanger of Example 2.15 with the open-loop transfer function e 5s G(s) = : (10s + 1)(60s + 1) (a) Design a lead compensator that yields PM possible closed-loop bandwidth.

45 and the maximum

(b) Design a PI compensator that yields PM possible closed-loop bandwidth.

45 and the maximum

Solution :

1

10

6181

Magnitude

Bode plot with Lead Com pens ation

0

10

-1

10

-1

0

10

10 ω (rad/s ec)

-100

Phase (deg)

-120 -140 -160 -180 -200 -1 10

0

10 ω (rad/s ec)

(a) First, make sure that the phase calculation includes the time delay lag of -Td ! = 5!: A convenient placement of the lead zero is at ! = 0:1 because that will preserve the -1 slope until the lead pole. We then raise the gain until the speci…ed PM is obtained in order to maximize the crossover frequency. The resulting lead compensator, D(s) =

90(s + 0:1) (s + 1)

yields P M = 46 as seen by the Bode below.Also note that the crossover frequency, ! c = 0:15 rad/sec, which can be read approximately from the plot above, and veri…ed by using the margin command in Matlab with the phase adjusted by the time delay lag. (b) The brealpoint of the PI compensator needs to be kept well below 0.1 in order to maintain a positive phase margin at as high a crossover frequency as possible. In Table 4.1, Zeigler-Nichols suggest a break-

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD point at ! = 1=17, so we will select a PI of the form :

D(s) = K 1 +

1 20s

and select the gain so that the PM speci…cation is met. For K = 0:55 the phase margin is 46 as shown by the Bode below:

Bode plot with PI com pens ation

1

Magnitude

10

0

10

-1

10

-2

10

10

-2

-1

10 ω (rad/s ec)

0

Phase (deg)

6182

-50 -100 -150 -200 -2 10

-1

10 ω (rad/s ec)

Note with this compensation that ! c = 0:02 rad/sec, which is considerably lower than that yielded by the lead compensation.

6183

Figure 6.105: Control system for Problem 67

Problems and Solutions for Section 6.9 67. A feedback control system is shown in Fig.6.105. The closed-loop system is speci…ed to have an overshoot of less than 30% to a step input. (a) Determine the corresponding PM speci…cation in the frequency domain and the corresponding closed-loop resonant peak value Mr . (See Fig. 6.38) (b) From Bode plots of the system, determine the maximum value of K that satis…es the PM speci…cation. (c) Plot the data from the Bode plots (adjusted by the K obtained in part (b)) on a copy of the Nichols chart in Fig. 6.84 and determine the resonant peak magnitude Mr . Compare that with the approximate value obtained in part (a). (d) Use the Nichols chart to determine the resonant peak frequency ! r and the closed-loop bandwidth. Solution : (a) From Fig. 6.38 : Mp

0:3 =) P M

40o =) Mr

resonant peak : Mr

1:5

1:5

(b) A sketch of the asymptotes of the open loop Bode shows that a PM of = 40o is obtained when K = 8: A Matlab plot of the Bode can be used to re…ne this and yields K = 7:81 for P M = 40o : (c) The Nichols chart below shows that Mr = 1:5 which agrees exactly with the prediction from Fig. 6.38:

6184

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

(d) The corresponding frequency where the curve is tangent to Mr = 1:5 is: ! r = 2:41 rad/sec as can be determined by noting the frequency from the Bode plot that corresponds to the point on the Nichols chart. The bandwidth ! BW is determined by where the curve crosses the closed-loop magnitude of 0.7 and noting the frequency from the Bode plot that corresponds to the point on the Nichols chart ! BW = 3:94 rad/sec 68. The Nichols plot of an uncompensated and a compensated system are shown in Fig. 6.106.

6185

Figure 6.106: Nichols plot for Problem 68

6186

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD (a) (b) (c) (d)

What What What What

are the are the are the type of

resonance peaks of each system? PM and GM of each system? bandwidths of each system? compensation is used?

Solution : (a) Resonant peak : Uncompensated system Compensated system

: :

Resonant peak = 1:5 (! r = 50 rad/sec) Resonant peak = 1::05 (! r = 20 rad/sec)

(b) PM, GM : Uncompensated system

:

Compensated system

:

1 =5 0:2 1 P M = 64 ; GM = = 10 0:1 P M = 42 ; GM =

(c) Bandwidth : Uncompensated system Compensated system

: :

Bandwidth = 70 rad/sec Bandwidth = 30 rad/sec

(d) Lag compensation is used, since the bandwidth is reduced.

6187 69. Consider the system shown in Fig. 6.97. (a) Construct an inverse Nyquist plot of [Y (j!)=E(j!)] 1 . (b) Show how the value of K for neutral stability can be read directly from the inverse Nyquist plot. (c) For K = 4, 2, and 1, determine the gain and phase margins. (d) Construct a root-locus plot for the system, and identify corresponding points in the two plots. To what damping ratios do the GM and PM of part (c) correspond? Solution : (a) See the inverse Nyquist plot.

(b) Let G(j!) =

Y (j!) E(j!)

The characteristic equation with s = j! : 1 + Ku G(j!) = 0 =) G

1

=

Ku

From the inverse Nyquist plot, Ku =

4 =) Ku = 4

(c) K 4 2 1

GM 4 4 =1 4 2 =2 4 1 =4

PM 0 18:3 38:1

6188

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

(d) K 4 2 1

closed-loop poles 4 j1:73 3:63 0:19 j1:27 3:37 0:31 j0:89

70. An unstable plant has the transfer function Y (s) s+1 = : F (s) (s 1)2

0 0:14 0:33

6189 A simple control loop is to be closed around it, in the same manner as the block diagram in Fig. 6.97. (a) Construct an inverse Nyquist plot of Y =F . (b) Choose a value of K to provide a PM of 45 . What is the corresponding GM? (c) What can you infer from your plot about the stability of the system when K < 0? (d) Construct a root-locus plot for the system, and identity corresponding points in the two plots. In this case, to what value of does PM = 45 correspond? Solution : (a) The plots are :

(b) From the inverse Nyquist plot, K = 3:86 provides a phase margin of 45 . Since K = 2 gives \G(j!) 1 = 180 , GM =

2 = 0:518 3:86

Note that GM is less than 1, but the system with K = 3:86 is stable. K = 3:86; GM = 0:518 (c) We can apply stability criteria to the inverse Nyquist plot as follows :

6190

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Let N P ( Z

= = = =

Net number of clockwise encirclement of K 1 Number of poles of G in RHP Number of zeros of G in RHP) Number of closed-loop system roots in RHP

Then, K< 2<

2

K1

K > 2 =) N = 0; P = 0 =) Z = 0 =) Stable 1 > K > 2 =) N = 2; P = 0 =) Z = 2 =) Two unstable closed-loop roots K < 1 =) N = 1; P = 0 =) Z = 1 =) One unstable closed-loop root

Then, we can infer from the inverse Nyquist plot the stability situation when K is negative. In summary, when K is negative, there are either one or two unstable roots, and the system is always unstable. (d) The stability situation seen in the root locus plot agrees with that obtained from the inverse Nyquist plot. They show : K>2 1 jp4 j > jp3 j: Thus, T2 = [t5 t1 t2 t4 t3 ]. (b) Our solution uses the Matlab sort command to re-order eigenvectors. Note that this approach is independent of the size of the system matrix Am . T is the same as in Problem 7.10. Because of the equality of the magnitudes of the complex eigenvectors, we can switch two of the columns of the matrix T2 . In Matlab, p = eig(F); [f, indices] = sort(abs(p)); T2 = T(:,indices); n = T2nG; T3 = T2*diag(n);

7015 Am2 = T3nF*T3; 2

6 6 Am2 = 6 6 4

0:0000 0:0000 0:0000 0:0000 0:0000

0:0000 0:5075 0:0000 0:0000 0:0000

0:0000 0:0000 0:6371 17:2941 0:0000

0:0123

2:4092 ,

0:0000 0:0000 0:0257 0:6371 0:0000

0:0000 0:0000 0:0000 0:0000 0:9683

3

7 7 7; 7 5

Bm2 = T3nG; T

Bm2 = [1 1 1 1 1] Cm2 = h3*T3; Cm2 =

2:8708

6:5602

1:2678

Dm2 = 0; 14. Find the characteristic equation for the modal-form matrix Am of Eq. (7.17a) using Eq. (7.58). Solution: det(sI

s+4 0 0 s+3

F) = det

= (s + 4)(s + 3)

Since Am was already in modal form, your solution is easily checked by inspection. 15. Given the system, 4 2

x_ =

1 1

0 1

x+

u;

with zero initial conditions, …nd the steady-state value of x for a step input u. Solution: We are given x_ = Fx + Gu. Steady-state means that x_ = 0 and a step input (or unit step) means u = 1(t). Thus, assuming that F is invertible (which you can check), we have, 0 = Fxss + G =) xss =

F

1

G=

4 2

1 1

1

0 1

=

1=6 2=3

:

This can be veri…ed in Matlab with step(F,G,H,J,1) where H=eye(size(F)) and J=[0;0]. 16. Consider the system shown in Fig. 7.83. G

S

1

5

H

2

1 s 1 4

x4

1 2s

G

2

5

H

1

1 5 s

x1

1

S 1

2

1 5 s

x2

S 1

x3

1

1

U

Figure 7.84: A block diagram for Problem 7.16.

Y

7016

CHAPTER 7. STATE-SPACE DESIGN a) Find the transfer function from U to Y . b) Write state equations for the system using the state variables indicated. Solution: (a) The system is equivalent to the block diagram shown. Following the block diagram back to known state variables, 1 ; y = x1 1 + 2s + s and, x1 =

1 u 2s(s + 4)

1 s

2s +

1 s

x1 ;

resulting in, Y (s) 2s3 + s2 + s = 4 : U (s) 2s + 8s3 + 2s2 + 1

Block diagram for solution of Problem 7.16 (a). Another possible solution is in terms of Mason’s rule. (b) 2

3 2 x_ 1 0 0 6 x_ 2 7 6 1 0 6 7 6 4 x_ 3 5 = 4 0 1 x_ 4 0 0 y

=

1

1

0 0 0 1 0

1

32

3 2 x1 0 6 x2 7 6 0 0 7 76 7+6 0 5 4 x3 5 4 0 x4 1 4 1 2

x:

3

7 7 u; 5

With practice, you should be able to see quickly that x_ 3 is simply the input to block H2 , which is the sum of x2 and x4 . But this is nothing more than the third row of the matrix equation above. Your results can be checked for consistency using Matlab’s command ss2tf. 17. Using the indicated state variables, write the state equations for each of the systems shown in Fig. 7.85. Find the transfer function for each system using both block-diagram manipulation and matrix algebra [as in Eq. (7.48)]. Solution: (a) Performing a partial fraction expansion on (s + 2)=(s + 4), Fig. 7.85(a) can be redrawn as shown below.

7017

U

1 s 1 4

x3

1

x2

s 1 2 s 1 3

1

S

x1

1

x1

1 s

1

S

Y

2 5

(a)

U

1

x4 1 s 1 10 1

1 s 1 2 x3

S 2

4 s

x2

x1 1 s 1 3 1

1

S

Y

(b)

Figure 7.85: Block diagrams for Problem 7.17.

Redrawn block diagram for solution to Problem 7.17(a). By inspection of the block diagram, the state equations are, 2 3 2 3 5 1 1 0 3 2 5 x + 4 1 5 u; x_ = 4 0 0 0 4 1 Computing G(s) = H(sI

y

=

F)

1

1

0

1

x:

G, the following transfer function is obtained: s2 + 10s + 20 Y (s) = 3 : U (s) s + 12s2 + 47s + 60

We can use the Matlab ss2tf command to verify this result. (b) Using the second block diagram given in 2 3 1 6 0 0 x_ = 6 4 0 1 0 0 y

=

1

0

1

Fig. 7.85(b), we can 3 2 0 0 0 6 0 4 0 7 7 x+ 6 4 1 2 1 5 0 10 1

0

x:

write, 3 7 7 u; 5

7018

CHAPTER 7. STATE-SPACE DESIGN Computing H(sI

F)

1

G (using Matlab’s ss2tf), we obtain the following transfer function, Y (s) s3 + 14s2 + 37s + 44 = 4 : U (s) s + 15s3 + 60s2 + 112s + 120

18. For each of the listed transfer functions, write the state equations in both control and observer canonical form. In each case draw a block diagram and give the appropriate expressions for F, G, and H. s2 2 (control of an inverted pendulum by a force on the cart) a) 2 2 s (s 1) 3s + 4 b) 2 s + 2s + 2 Solution: (a) Y (s) s2 2 = 2 2 : U (s) s (s 1) This transfer function can be realized in controller canonical form as shown below. From the …gure, we have, 2

3 2 x_ 1 0 6 x_ 2 7 6 6 7 = 6 1 4 x_ 3 5 4 0 0 x_ 4 y

=

0

1 0 1 0

0 0 0 1

1

0

32 x1 0 6 0 7 7 6 x2 0 5 4 x3 0 x4 2

x:

3 1 7 6 0 7 7 + 6 7 u; 5 4 0 5 0 3

2

Controller canonical form for the transfer function of Problem 7.18(a). The block diagram for observer canonical form is shown below. From the …gure, we have, 2

3 2 x_ 1 0 6 x_ 2 7 6 1 6 7 6 4 x_ 3 5 = 4 0 x_ 4 0 y

=

1

1 0 0 0

0 1 0 0

0

0

32 0 x1 6 x2 0 7 76 1 5 4 x3 0 x4

0

x:

3

2

3 0 7 6 1 7 7+6 7 5 4 0 5 u; 2

7019

Observer canonical form for the transfer function of Problem 7.18(a).

(b) Y (s) 3s + 4 = 2 : U (s) s + 2s + 2

This transfer function can be realized in Controller canonical form as shown below. From the …gure, we have, x_ 1 x_ 2

2 1

= y

=

3

2 0 4

x1 x2

+

1 0

u;

x:

Controller canonical form for the transfer function of Problem 7.18(b).

The block diagram for observer canonical form is shown below. From the …gure, we have: x_ 1 x_ 2

2 2

= y

=

1

1 0 0

x1 x2 x:

+

3 4

u;

7020

CHAPTER 7. STATE-SPACE DESIGN

1 8:pdf

Observer canonical form for the transfer function of Problem 7.18(b). (s) 19. Consider the transfer function, G(s) = YU (s) = a) By rewriting Eq. (7.263) in the form,

G(s) =

s+1 s2 +5s+6 :

1 s+3

s+1 s+2

(7.263)

;

…nd a series realization of G(s) as a cascade of two …rst-order systems. b) Using a partial-fraction expansion of G(s), …nd a parallel realization of G(s). c) Realize G(s) in control canonical form. Solution: (a) The series realization shown below is given by: G(s) =

1 s+3

s+1 s+2

= g^2 (s)^ g1 (s):

Series connection of G(s) for Problem 7.19(a).

For g^1 (s); x_ 1 For g^2 (s); x_ 2

= =

2x1 + u1 ; y1 = x1 + u1 : 3x2 + u2 ; y2 = x2 :

The series interconnections result in u = u1 ; y = y2 ; u2 = y1 . Therefore, x_ 1 x_ 2

2 1

= y

=

0

0 3 1

x:

x1 x2

+

1 1

u;

7021 The Matlab command series can also be used. (b) The parallel realization, shown below, is given by: G(s) =

1 2 + = g^1 (s) + g^2 (s): s+3 s+2

Parallel connection of G(s) for Problem 7.19(b). For g^1 (s); For g^2 (s);

x_ 1 x_ 2

= =

3x1 + u1 ; y1 = 2x1 : 2x2 + u2 ; y2 = x2 :

The interconnections are u1 = u2 = u; y = y1 + y2 . Therefore, x_ 1 x_ 2

3 0

= y

2

=

0 2 1

x1 x2

+

1 1

u;

x:

The Matlab command parallel can also be used. (c) Control canonical form, shown in Fig. 7.17, is realized by simply picking o¤ the appropriate coe¢ cients of the original (strictly proper) transfer function. If the original function is not strictly proper, then it should be reduced to a feedthrough term plus a strictly proper term.

Controller canonical form of G(s) for Problem 7.19(c). For G(s), we have, F=

5 1

6 0

; G=

1 0

; H=

1

1

:

Problems and Solutions for Section 7.5: Control-Law Design for Full-State Feedback

7022

CHAPTER 7. STATE-SPACE DESIGN

20. Consider the plant described by, 0 7

x_ =

1 4

y=[ 1

x+

1 2

u;

3 ]x:

a) Draw a block diagram for the plant with one integrator for each state variable. b) Find the transfer function using matrix algebra. c) Find the closed-loop characteristic equation if the feedback is (1) u =

[ K1

K2 ]x; (2) u =

Ky.

Solution:

State realization showing integrators explicitly. (a) See …gure. (b) Using the formula G(s) = H(sI

F)

G(s) =

1

G, we obtain,

7s + 27 Y (s) = 2 : U (s) s + 4s 7

The Matlab command ss2tf can also be used. (c) (i) State feedback, u = det( I

[K1 K2 ]x.

F + GK)

=

det

=

2

+ K1 7 + 2K2

1 + K2 + 4 + 2K2

+ (4 + 2K2 + K1 ) + (6K1 + 7K2

(ii) Output feedback, u=

Ky =

K

1

3

x=

K

3K

This yields the following closed-loop characteristic equation: 2

+ (7K + 4) + (27K

7) = 0:

x:

7) = 0:

7023 Hints: If you have already solved the case for state feedback, simply plug K1 = K and K2 = 3K into the characteristic equation for state feedback and …nd the characteristic equation for output feedback. The output vector H …xes the ratio among the state variables. Secondly, although there were products of K1 and K2 when we were forming the determinant, they should all cancel in your …nal answer. The reason for this is that the characteristic equation x_ = (F GK)x is linear in K. 21. For the system, 0 6

x_ = y

=

1

1 5 0

x+

0 1

u;

x;

design a state feedback controller that satis…es the following speci…cations: Closed-loop poles have a damping coe¢ cient

= 0:707.

Step-response peak time is under 3.14 sec. Verify your design with Matlab. Solution: For a second-order system, the speci…cation on rise time canpbe translated into a value of ! n by 2 . This yields ! n = 1:414. the equation ! d = . Then determine ! n from ! d = ! n 1 Using full state feedback, we would like the a characteristic equation to be, s2 + 2 ! n s + ! 2n = s2 + 2s + 2 = 0: Using state feedback u =

Kx, we get, x_ = (F

GK)x =

0 6

1 k1

5

k2

x:

Hence the closed-loop characteristic equation is, s2 + (5 + k2 )s + (6 + k1 ) = 0:

Comparing coe¢ cients, k1 = 4 and k2 = 3. The Matlab command place can also be used. The reference step can be simulated in Matlab with u = Kx + r, and the Matlab command step, as shown below.

7024

CHAPTER 7. STATE-SPACE DESIGN

Step Response 0.7

0.6

Amplitude

0.5

0.4

0.3

0.2

0.1

0 0

1

2

3 Time (sec)

4

5

6

Step response for Problem 7.21. 22. a) Design a state feedback controller for the following system so that the closed-loop step response has an overshoot of less than 25% and a 1% settling time under 0.115 sec.: x_ =

0 0

y

1

=

1 10 0

x+

0 1

u;

x:

b) Use the step command in Matlab to verify that your design meets the speci…cations. If it does not, modify your feedback gains accordingly. Solution: (a) For the overshoot speci…cation, Mp = e

p

1

2

< 25% =)

= 0:4:

For the 1% settling time speci…cation, we use, e

! n ts

= 0:01 =) ! n =

4:6 : ts

7025 (b) This can be implemented in Matlab with the following code: F = [0,1;0,-10]; G = [0;1]; H = [1,0]; J = 0; zeta = 0.404; % Tweak values slightly so that specs are met. ts = 0.114; wn = 4.6/(ts*zeta); p = roots([1, 2*zeta*wn, wn^2]); k = place(F,G,p); sysCL=ss(F-G*k,G,H,J) step(sysCL); The step response is shown next.

1.4

x 10

-4

Control des ign with overs hoot and s ettling tim e s pecs

1.2

1

y(t)

0.8

0.6

0.4

0.2

0

0

0.02

0.04

0.06 0.08 Tim e (s ec)

Step response for Problem 7.22.

0.1

0.12

0.14

7026

CHAPTER 7. STATE-SPACE DESIGN

23. Consider the system, 2

x_ = 4 y

=

1 0 1 1

0

2 1 0 0

x:

3 2 3 2 2 1 5 x + 4 0 5 u; 1 1

a) Design a state feedback controller for the system so that the closed-loop step response has an overshoot of less than 5% and a 1% settling time under 4.6 sec. b) Use the step command in Matlab to verify that your design meets the speci…cations. If it does not, modify your feedback gains accordingly. Solution: (a) There are many di¤erent approaches to designing the control law. We will attack the problem using a symmetric root locus. We assume the output is x1 . Although the system is third-order, we can still use the second-order order rules of thumb in order to get an estimate of where we would like the closed loop poles. = Mp

4:6 =) ! n = 1; ts 5% =) > 0:7:

The open-loop poles are at 1:45 and 0:77 j1:47 and the open-loop zeros are at 1:37 and 0:37. The symmetric root locus is shown on the next page and was generated using the following Matlab code: %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % the function srl is used to compute the roots of the symmetric root locus function [k,p]=srl(f,g,h) a=[f 0*f;-h’*h -f’]; b=[g;0*g]; c=[0*h g’]; rlocus(a,b,c,0); [k,p]=rloc…nd(a,b,c,0) Note that crosses indicate where the closed-loop pole locations have been selected, which roughly correspond to the and ! n suggested by the rules of thumb for a second-order system with no zeros. The control gains K = 0:78 0:07 0:28 correspond to these closed loop pole locations. The Matlab command place can be used to verify this computation. The step response is shown next using the Matlab step command. Technically, this closed-loop step response meets the 4:6 sec 1% settling time and 5% overshoot. However, the right half plane zero close to the origin gives catastrophic results in terms of undershoot. This should alert the reader to the importance of paying attention to the zeros of the system, especially in the right half plane.

7027

Root Locus 2

1.5

Imaginary Axis

1

0.5

0

-0.5

-1

-1.5

-2 -4

-3

-2

-1

0 Real Axis

1

Symmetric root locus for Problem 7.23.

2

3

4

7028

CHAPTER 7. STATE-SPACE DESIGN

Control des ign with overs hoot and s ettling tim e s pecs 1 0.8 0.6 0.4

y(t)

0.2 0 -0.2 -0.4 -0.6 -0.8 -1

0

0.5

1

1.5

2 2.5 Tim e (s ec)

3

3.5

4

4.5

Closed-loop step response for Problem 7.23. 24. Consider the system in Fig. 7.83. U

s s2 1 4

Y

Figure 7.86: System for Problem 7.24.

a) Write a set of equations that describes this system in the control canonical form as x_ = Fx + Gu and y = Hx. b) Design a control law of the form, u=

[ K1

that will place the closed-loop poles at s =

K2 ] 2

x1 x2

2j:

Solution: (a) Let’s write this system in the control canonical form,

;

7029

x_ 1 x_ 2 y (b) If u =

=

0 1

=

1

4 0 0

x1 x2

+

1 0

u;

x:

[K1 K2 ]x, the poles of the closed-loop system satisfy det(sI det

s + K1 4

1 + K2 s

F + GK) = 0. Thus,

= 0 =) s2 + K1 s + 4 + K2 = 0:

The closed-loop characteristic equation is, (s + 2

2j)(s + 2 + 2j) = s2 + 4s + 8 = 0:

Comparing coe¢ cients, we have K1 = 4 and K2 = 4. The Matlab command place can also be used to verify this result. 25. Output Controllability: In many situations a control engineer may be interested in controlling the output y rather than the state x. A system is said to be output controllable if at any time you are able to transfer the output from zero to any desired output y in a …nite time using an appropriate control signal u . Derive necessary and su¢ cient conditions for a continuous system (F,G,H) to be output controllable. Are output and state controllability related? If so, how? Solution: Because we are considering linear systems, if you can take the state from some initial state to some …nal condition in a …nite time with a …nite input, then you can also take it to the same state in in…nitesimal time using impulsive inputs. To express this mathematically, let u be, u(t) = g1 (t) + g2

(1)

(t) +

+ gn

(n 1)

(t);

where (t) represents a delta function, (1) (t) represents the …rst derivative of a delta function (a unit doublet), etc., and the gi are scalars to be determined. Let, u = [g1 g2

gn ]T ;

then x(0+)

x(0 ) = C u :

Hence, we have found a control signal that will drive the state to arbitrary values given the non-singularity of the controllability matrix, C.

In fact, the invertibility of C is a necessary and su¢ cient condition for state controllability. For output controllability, Hx(0+) Hx(0 ) = HC u ; y(0+) y(0 ) = HC u : Assuming (without loss of generality) that y(0 ) = 0, we have, y(0+) = [HG HFG

HFn

1

G]u :

7030

CHAPTER 7. STATE-SPACE DESIGN Therefore, a system is output controllable if and only if, HFn

[HG HFG

1

G] is full rank:

This is always true (for a single-input single-output system) unless the transfer function is zero. Of course, state controllability implies output controllability, but output controllability does not imply state controllability. 26. Consider the system,

2

6 x_ = 6 4

0 1 5 0

4 0 4 0 7 1 0 3

3 2 0 6 0 7 7x + 6 4 15 5 3

3 0 0 7 7 u: 1 5 0

a) Find the eigenvalues of this system. (Hint: Note the block-triangular structure.) b) Find the controllable and uncontrollable modes of this system. c) For each of the uncontrollable modes, …nd a vector v such that, vT G = 0;

vT F = vT :

d) Show that there are an in…nite number of feedback gains K that will relocate the modes of the system to 5, 3, 2, and 2. e) Find the unique matrix K that achieves these pole locations and prevents initial conditions on the uncontrollable part of the system from ever a¤ecting the controllable part. Solution: (a) Because the system is block lower triangular, we can determine the eigenvalues of the system by taking the union of the eigenvalues of each of the blocks along the main (block)diagonal. s2 + 4s + 4 = 0 =) 2; 2: s2 + 2s 48 = 0 =) 6; 8 Thus, the eigenvalues of the system are command).

2; 2; 6, and

8. (Easily checked with Matlab’s eig

(b) To …nd the controllable or uncontrollable modes of the system, we follow method learned in Problem 7.28. Speci…cally, we …nd an orthogonal similarity transformation which transforms (F, G, H) to (F, G, H) where F is an upper-Hessenberg matrix. (See Problem 7.28 for details). Observe that this system is almost in the desired form already! Simply by interchanging the state variables x3 and x4 , we can transform the system into the proper form. 2

3 2 x_ 1 0 6 x_ 2 7 6 1 6 7 6 4 x_ 4 5 = 4 0 x_ 3 5

4 4 0 7

32 0 0 x1 6 x2 0 0 7 76 3 3 5 4 x4 15 1 x3

3

2

3 0 7 6 0 7 7 + 6 7 u: 5 4 0 5 1

Now the controllable and uncontrollable modes can be determined by inspection. The uncontrollable modes correspond to the eigenvalues of the F11 block, so 2 and 2 are both uncontrollable modes. Similarly, the controllable modes from the F22 block are 8 and 6.

7031

a k l

u

u

1

u

2

u

m

m

Figure 7.87: Coupled pendulums for Problem 7.27.

Matlab function ctrbf will give a similar result, although the order of the state variables may be switched. (c) Notice that we need the left eigenvectors of F that is orthogonal to G. The only left T eigenvector of F that is orthogonal to G is 1 2 0 0 : (d) Because the modes at 2 and 2 are uncontrollable, we expect that state feedback will not have any a¤ect on these modes. Writing an expression for the feedback we have, 2 32 3 3 2 x_ 1 x1 0 4 0 0 6 x_ 2 7 6 7 6 x2 7 1 4 0 0 6 76 7 7 6 4 x_ 3 5 = 4 5 k1 7 k2 1 k3 15 k4 5 4 x3 5 : x4 x_ 4 0 0 3 3 Notice that the system matrix is still block diagonal. The characteristic equation of the F22 block gives, det(sI F22 ) = s2 + (2 + k3 )s + (3k3 + 3k4 48) = 0: Picking k3 = 6 and k4 = 15 will place the controllable roots at 3 and 5. Since k1 and k2 are arbitrary, there are an in…nite number of feedback gains that will relocate the modes of the system to the desired locations. (e) To completely decouple the controllable and uncontrollable portions of the system, we make the F21 block identically zero by setting k1 = 5 and k2 = 7. 27. Two pendulums, coupled by a spring, are to be controlled by two equal and opposite forces u, which are applied to the pendulum bobs as shown in Fig. 7.86. The equations of motion are ml2 •1 = ml2 •2 =

ka2 ( 2

ka (

1

2)

mgl

1

lu;

2

1)

mgl

2

+ lu:

a) Show that the system is uncontrollable. Can you associate a physical meaning with the controllable and uncontrollable modes? b) Is there any way that the system can be made controllable?

7032

CHAPTER 7. STATE-SPACE DESIGN Solution: (a) Using the state vector x = [ 1 _ 1 2 0 6 g ka2 6 ml2 + l 6 x_ = 6 0 4 2

_ 2 ]T ;

2

ka ml2

1

0

0 0

ka2 ml2

0 ka2 ml2

0

+

3

2 7 0 7 6 7x + 6 7 4 1 5 0

0

g l

0 1 ml

0 1 ml

3

7 7 u: 5

The controllability matrix is determined as, C

=

2

6 6 6 = 6 6 6 4

G

F2 G

FG 0

1 ml

1 ml

0

F3 G ka2 ml2

1 ml

0

1 ml

1 ml

0

+

g l

+

ka2 m3 l3

1 ml

ka2 m3 l3 ka2 ml2

+

+

g l

+

3

ka2 m3 l3

7 7 7 7 7 7 5

0

0 ka2 m3 l3

ka2 ml2

1 ml

0

g l

1 ml

ka2 ml2

+

g l

0

Then (F,G) is uncontrollable since det(C)=0. If we re-write the state equations in terms of the state vector, _ T; z= _ where,

=

1

+

2;

and,

=

1

ml2 • ml2 •

2

, then the resulting equations of motion are, = =

mgl 2ka2

mgl

2lu:

Clearly, , the “pendulum mode” (or the symmetric mode, i.e., the pendulums swinging together), is uncontrollable and, , the “spring mode”(i.e., the unsymmetric mode) is controllable. (b) Yes, make the forces unequal, i.e., let u1 6= u2 ; or eliminate one of the forces, i.e., let u1 = 0; or let u2 = 0: 28. The state-space model for a certain application has been description matrices: 2 3 2 0:174 0 0 0 0 0:207 6 0:157 0:645 0 0 0 7 6 0:005 6 7 6 6 1 0 0 0 7 0 F=6 6 0 7; G = 6 4 0 5 4 0 1 0 0 0 0 0 0 1 0 0

given to us with the following state 3

7 7 7; 7 5

H=[ 1

0

0

0

0 ]:

a) Draw a block diagram of the realization with an integrator for each state variable. b) A student has computed det C = 2:3 10 7 and claims that the system is uncontrollable. Is the student right or wrong? Why? c) Is the realization observable? Solution: (a) The block diagram is shown below.

7033

Block diagram for Problem 7.28. (b) The system is controllable because a control signal u (command) reaches all the state variables of the system through the integrators in Fig. 7.26. The determinant of the controllability matrix is small, det(C) = 2:3 10 7 , due to poor scaling of system variables. For example, if the control signal is scaled by 100, then det(C) = 2:3 103 . (c) The realization is unobservable. You can check det(O) or just observe from the block diagram that there is no path from the state variables x2 ; x3 ; x4 , or x5 to the output y. 29. Staircase Algorithm (Van Dooren et al., 1978): Any realization (F,G,H) can be transformed by an orthogonal similarity transformation to (F,G,H), where F is an upper Hessenberg matrix (having one nonzero diagonal above the main diagonal): 2

6 6 F = T FT = 6 6 4 T

where g1 6= 0, and,

1

0 .. . ..

.

H = HT = [h1

0 0 n 1

hn ];

3

7 7 7; 7 5 T

2

0 0 .. .

6 6 6 G=T G=6 6 4 0 g1 T

1

3

7 7 7 7; 7 5

= TT :

Orthogonal transformations correspond to a rotation of the vectors (represented by the matrix columns) being transformed with no change in length. a) Prove that if i = 0 and i+1 ; : : : ; n 1 6= 0 for some i, then the controllable and uncontrollable modes of the system can be identi…ed after this transformation has been done. b) How would you use this technique to identify the observable and unobservable modes of (F, G, H)? c) What advantage does this approach for determining the controllable and uncontrollable modes have over transforming the system to any other form? d) How can we use this approach to determine a basis for the controllable and uncontrollable subspaces, as in Problem 7.43? This algorithm can be used to design a numerically stable algorithm for pole placement [see Minimis and Paige (1982)]. The name of the algorithm comes from the multi-input version in which the i are the blocks that make F resemble a staircase. Solution:

7034

CHAPTER 7. STATE-SPACE DESIGN (a) If

i

= 0, 2

1

6 6 6 6 6 F = TT FT 6 6 6 6 4

0 2

0 0 0

0 0 0 i+1

0 0 0 0 .. .

0 0 0 0

3

0 0 0 0

0

7 7 7 7 7 7= 7 7 7 5

0 n 1

2

F11 F21

0 F22

;

6 6 6 6 T G=T G=6 6 6 6 4

0 0 0 0 0 0 g1

3

7 7 7 7 7; 7 7 7 5

This suggests naturally splitting up the state vector into two parts x = [x1 x2 ]T where x1 and x2 are vectors of the appropriate size (depending upon which i = 0). Then recognize that the equations are, x_ 1 x_ 2

= F11 x1 ; = F21 x1 + F22 x2 + g1 u:

Notice that the control signal u and the state x2 do not e¤ect the state x1 . Thus, all of the modes associated with the block F11 are uncontrollable. All of the states in x2 are controllable. This is easily checked by forming the controllability matrix associated with the pair (F22 ,g1 ). Hence, the system has been split into its controllable and uncontrollable parts. (b) Use duality, i.e., transform [FT HT ] into Hessenberg form. (c) Because T

1

= TT , three advantages are recognized:

(i) Better numerical accuracy. (ii) The controllable-uncontrollable decomposition is immediate. (iii) Repeated roots are handled. (d) Simply split T and extract the controllable and uncontrollable subspaces, T =

[ T1 T2 ]; |{z} |{z} i

T1

(3)

n i

T

= N (C );

T2 = R(C):

(4)

See the Matlab ctrbf (and obsvf) functions.

Problems and Solutions for Section 7.6: Selection of Pole Locations for Good Design 30. The normalized equations of motion for an inverted pendulum at angle • = + u;

x •=

on a cart are,

u;

where x is the cart position, and the control input u is a force acting on the cart. a) With the state de…ned as x = [ ; _ ; x; x] _ T , …nd the feedback gain K that places the closedloop poles at s = 1; 1; 1 1j. For parts (b) through (d), assume that = 0:5. b) Use the symmetric root locus to select poles with a bandwidth as close as possible to those of

7035 part (a), and …nd the control law that will place the closed-loop poles at the points you selected. c) Compare the responses of the closed-loop systems in parts (a) and (b) to an initial condition of = 10 . You may wish to use the initial command in Matlab. d) Compute Nx and Nu for zero steady-state error to a constant command input on the cart position, and compare the step responses of each of the two closed-loop systems. Solution: (a) The state space equations of motion are, 2

3 2 _ 0 6 • 7 6 1 6 7 6 4 x_ 5 = 4 0 x •

32 1 0 0 6 _ 0 0 0 7 76 5 0 0 1 4 x 0 0 0 x_

We require the closed-loop characteristic equation to be, c (s)

3 0 7 6 1 7 7 7+6 5 4 0 5 u: 1 3

2

= (s + 1)2 (s2 + 2s + 2) = s4 + 4s3 + 7s2 + 6s + 2:

From the above state equations, det(sI

F + GK) = s4 + (k2

k4 )s3 + (k1

k3

1)s2 + k4 (1

)s + k3 (1

)

c (s)

Comparing coe¢ cients yields: k1

=

K

=

10 8 10 4 2 ; k2 = ; k3 = 1 1 1 12 16 4 12 :

; k4 =

6 1

;

(b) The symmetric root locus is shown below, where we have chosen H = 0 0 1 following Matlab commands can be used to generate the symmetric root locus,

0 . The

% Symmetric root locus a=[F, 0*F;-H’*H, -F’]; b=[G;0*G]; c=[0*H, G’]; d=0; rlocus(a,b,c,d); The chosen pole locations, shown on the symmetric root locus, result in a feedback gain of (using Matlab’s place command), K=

13:5

18:36

3:9

13:98

:

(c) The initial condition response to (0) = 10 for both control designs in (a) and (b) is shown on the next page. (d) To compute Nx and Nu for zero steady-state error to a constant command input on cart position, x, we solve the equations, F H2

G J

Nx Nu

=

0 1

:

7036

CHAPTER 7. STATE-SPACE DESIGN This yields Nx = [0 0 1 0]T and Nu = 0. The step responses for each of the closed-loop systems (using the Matlab step command) are shown next.

Root Locus 3

2

Imaginary Axis

1

0

-1

-2

-3 -3

-2

-1

0 Real Axis

1

Symmetric root locus for Problem for Problem 7.30.

2

3

7037

Response to Initial Conditions 20

Pole placem ent SRL

Amplitude

15

10

5

0

-5 0

1

2

3

4 5 Time (sec)

6

7

Initial condition response with (0) = 10 for Problem 7.30.

8

9

7038

CHAPTER 7. STATE-SPACE DESIGN

1.2 1 0.8 0.6 0.4 0.2 0 -0.2

SRL Pole placem ent

-0.4 -0.6

0

1

2

3

4

5

6

7

8

9

Step response using Nx and Nu for Problem 7.30. 31. Consider the feedback system in Fig. 7.88. Find the relationship between K, T , and such that the closed-loop transfer function minimizes the integral of the time multiplied by the absolute value of the error (ITAE) criterion, Z J =

1

tjejdt;

0

for a step input. Assume ! 0 = 1. U

1

S

K 1 1 Ts

1 s2 1 2j s 1 1

Y

2

Figure 7.88: Control system for Problem 31.

Solution: From the diagram: Y (s) K=T = U (s) s3 + ( T1 + 2 )s2 + (2 +

2 T

)s +

2+K T

:

7039 From the ITAE requirements [see Franklin, Powell, Emami-Naeini 3rd. Edition, pp. 508], we need to have, c (s)

= s3 + 1:75s2 + 2:15s + 1:

Comparing the coe¢ cients, 1 + 2 = 1:75; T

2+

2 = 2:15; T

2+K = 1: T

32. Prove that the Nyquist plot for LQR design avoids a circle of radius one centered at the -1 point as shown in Fig. 7.89. Show that this implies that 12